Analytical Geometry - Conics - Tangents and Normals

Find the equation of the chord of contact of tangents from the point $\left(5,3\right)$ to the hyperbola $4x^2 - 6y^2 = 24$.

Given equation of hyperbola is $\;$ $4x^2 - 6y^2 = 24$

i.e. $\;$ $\dfrac{x^2}{24 / 4} - \dfrac{y^2}{24 / 6} = 1$

i.e. $\;$ $\dfrac{x^2}{6} - \dfrac{y^2}{4} = 1$

The transverse axis of the given hyperbola is along the X axis.

Comparing with the standard equation of the hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 6$ $\;$ and $\;$ $b^2 = 4$

Equation of chord of contact of tangents from the point $\left(x_1, y_1\right)$ to the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is: $\;$ $\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1$

Here $\;$ $\left(x_1, y_1\right) = \left(5,3\right)$

$\therefore$ $\;$ The required equation is

$\dfrac{5x}{6} - \dfrac{3y}{4} = 1$

i.e. $\;$ $10x - 9y - 12 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the chord of contact of tangents from the point $\left(-3,1\right)$ to the parabola $y^2 = 8x$.

Equation of parabola is $\;$ $y^2 = 8x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 8 \implies a = 2$

Equation of the chord of contact of tangents from the point $\left(x_1, y_1\right)$ to the parabola $y^2 = 4ax$ is

$yy_1 = 2a \left(x + x_1\right)$

Here, $\left(x_1, y_1\right) = \left(-3,1\right)$

$\therefore$ $\;$ Required equation is

$1 \times y = 2 \times 2 \times \left(x - 3\right)$

i.e. $\;$ $4x - y - 12 = 0$

Analytical Geometry - Conics - Tangents and Normals

Show that the line $x - y + 4 = 0$ is a tangent to the ellipse $x^2 + 3y^2 = 12$. Find the coordinates of the point of contact.

Equation of given line is $\;$ $x - y + 4 = 0$

i.e. $\;$ $y = x + 4$ $\;\;\; \cdots \; (1)$

Slope of given line $= m = 1$

Intercept of given line $= c = 4$

Equation of ellipse is $\;$ $x^2 + 3y^2 = 12$

i.e. $\;$ $\dfrac{x^2}{12} + \dfrac{y^2}{12 / 3} = 1$

i.e. $\;$ $\dfrac{x^2}{12} + \dfrac{y^2}{4} = 1$ $\;\;\; \cdots \; (2)$

The major axis of the ellipse given by equation $(2)$ is along the X axis.

$\therefore$ $\;$ Comparing equation $(2)$ with the tandard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 12$ $\;$ and $\;$ $b^2 = 4$

Condition that the given line is a tangent to the ellipse is $\;$ $c^2 = a^2 m^2 + b^2$ $\;\;\; \cdots \; (3)$

Substituting the values of $c$, $a^2$, $m$ and $b^2$ in equation $(3)$ we have,

$\left(4\right)^2 = 12 \times \left(1\right)^2 + 4$

i.e. $\;$ $16 = 16$ $\;$ which is true.

$\therefore$ $\;$ The line given by equation $(1)$ is a tangent to the ellipse given by equation $(2)$.

The point of contact of the tangent with the ellipse is $\;$ $\left(\dfrac{-a^2 m}{c}, \dfrac{b^2}{c}\right)$

i.e. $\;$ $\left(\dfrac{-12 \times 1}{4}, \dfrac{4}{4}\right)$ $\;\;$ i.e. $\;$ $\left(-3, 1\right)$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the two tangents that can be drawn from the point $\left(1, 2\right)$ to the hyperbola $2x^2 - 3y^2 = 6$.

Equation of hyperbola is $\;$ $2x^2 - 3y^2 = 6$

i.e. $\;$ $\dfrac{x^2}{6 / 2} - \dfrac{y^2}{6 / 3} = 1$

i.e. $\;$ $\dfrac{x^2}{3} - \dfrac{y^2}{2} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis of the hyperbola given by equation $(1)$ is along the X axis.

$\therefore$ $\;$ Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 3$ $\;$ and $b^2 = 2$

Let the equation of the tangent be $\;$ $y = mx + \sqrt{a^2 m^2 - b^2}$ $\;\;\; \cdots \; (2)$

Substituting the values of $a^2$ and $b^2$ in equation $(2)$, we have,

$y = mx + \sqrt{3m^2 - 2}$ $\;\;\; \cdots \; (3)$

Equation $(3)$ passes through the point $\left(1, 2\right)$.

$\therefore$ $\;$ We have from equation $(3)$,

$2 = m + \sqrt{3m^2 - 2}$

i.e. $\;$ $2 - m = \sqrt{3m^2 - 2}$ $\;\;\; \cdots \; (4)$

Squaring both sides of equation $(4)$ we get,

$4 + m^2 - 4m = 3m^2 - 2$

i.e. $\;$ $2m^2 + 4m - 6 = 0$

i.e. $\;$ $m^2 + 2m - 3 = 0$

i.e. $\;$ $\left(m + 3\right) \left(m - 1\right) = 0$

i.e. $\;$ $m = -3$ $\;$ or $\;$ $m = 1$

Substituting $m = -3$ in equation $(3)$ gives

$y = -3x + \sqrt{3 \times \left(-3\right)^2 - 2}$

i.e. $\;$ $3x + y - 5 = 0$ $\;\;\; \cdots \; (5a)$

Substituting $m = 1$ in equation $(3)$ gives

$y = x + \sqrt{3 \times \left(1\right)^2 - 2}$

i.e. $\;$ $x - y + 1 = 0$ $\;\;\; \cdots \; (5b)$

Equations $(5a)$ and $(5b)$ are the required equations of tangents.

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the two tangents that can be drawn from the point $\left(2,-3\right)$ to the parabola $y^2 = 4x$.

Equation of parabola is $\;$ $y^2 = 4x$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives $\;$ $a = 1$

Let the equation of tangent be $\;$ $y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (2)$

Substituting the value of $a$ in equation $(2)$ gives

$y = mx + \dfrac{1}{m}$ $\;\;\; \cdots \; (3)$

Equation $(3)$ passes through the point $\left(2, -3\right)$.

$\therefore$ $\;$ We have $\;$ $-3 = 2m + \dfrac{1}{m}$

i.e. $\;$ $2m^2 + 3m + 1 = 0$

i.e. $\;$ $\left(2m + 1\right) \left(m + 1\right) = 0$

i.e. $\;$ $m = -1$ $\;$ or $\;$ $m = \dfrac{-1}{2}$

When $m = -1$, we have from equation $(3)$, $\;$ $y = -x -1$

i.e. $\;$ $x + y + 1 = 0$ $\;\;\; \cdots \; (4a)$

When $m = \dfrac{-1}{2}$, we have from equation $(3)$, $\;$ $y = \dfrac{-x}{2} - \dfrac{1}{1 / 2}$

i.e. $\;$ $y = \dfrac{-x}{2} - 2$

i.e. $\;$ $x + 2y + 4 = 0$ $\;\;\; \cdots \; (4b)$

Equations $(4a)$ and $(4b)$ are the required equations of tangents.

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangents to the hyperbola $4x^2 - y^2 = 64$, which are parallel to $10x-3y+9=0$.

Equation of hyperbola is $\;$ $4x^2 - y^2 = 64$

i.e. $\;$ $\dfrac{x^2}{64 / 4} - \dfrac{y^2}{64} = 1$

i.e. $\;$ $\dfrac{x^2}{16} - \dfrac{y^2}{64} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis of the hyperbola given by equation $(1)$ is along the X axis.

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 16 \implies a = 4$ $\;$ and $\;$ $b^2 = 64 \implies b = 8$

Equation of given line is $\;$ $10x - 3y + 9 = 0$

i.e. $\;$ $y = \dfrac{10}{3} x + 3$

Slope of given line $= m = \dfrac{10}{3}$

$\because$ $\;$ the required tangents are parallel to the given line,

$\therefore$ $\;$ slope of required tangents $= m = \dfrac{10}{3}$

Equations of tangents (with slope $m$) to the hyperbola are $\;$ $y = mx \pm \sqrt{a^2 m^2 - b^2}$

$\therefore$ $\;$ Required equations of tangents to the hyperbola are

$y = \dfrac{10}{3} x \pm \sqrt{16 \times \dfrac{100}{9} - 64}$

i.e. $\;$ $y = \dfrac{10}{3} x \pm \sqrt{\dfrac{1024}{9}}$

i.e. $\;$ $y = \dfrac{10}{3} x \pm \dfrac{32}{3}$

i.e. $\;$ $10 x - 3y \pm 32 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangents to the ellipse $\dfrac{x^2}{20}+ \dfrac{y^2}{5} = 1$, which are perpendicular to $x+y+2=0$.

Equation of ellipse is $\;$ $\dfrac{x^2}{20} + \dfrac{y^2}{5} = 1$ $\;\;\; \cdots \; (1)$

The major axis of the ellipse given by equation $(1)$ is along the X axis.

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 20$ $\;$ and $\;$ $b^2 = 5$

Equation of given line is $\;$ $x + y + 2 = 0$

i.e. $\;$ $y = -x - 2$

Slope of given line $= m_1 = -1$

$\because$ $\;$ the required tangents are perpendicular to the given line,

slope of tangent $= m = \dfrac{-1}{m_1} = 1$

Equations of tangents (with slope $m$) to the ellipse are $\;$ $y = mx \pm \sqrt{a^2 m^2 + b^2}$

$\therefore$ $\;$ Required equations of tangents to the ellipse are

$y = 1 \times x \pm \sqrt{20 \times 1^2 + 5}$

i.e. $\;$ $y = x \pm 5$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangent to the parabola $y^2 = 6x$, parallel to $3x - 2y + 5 = 0$.

Equation of parabola is $\;$ $y^2 = 6x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 6 \implies a = \dfrac{3}{2}$

Equation of given line is $\;$ $3x - 2y + 5 = 0$

i.e. $\;$ $y = \dfrac{3}{2}x + \dfrac{5}{2}$

$\therefore$ $\;$ Slope of the given line $= m = \dfrac{3}{2}$

$\because$ $\;$ the tangent to the parabola is is parallel to the given line, slope of tangent to the parabola $= m = \dfrac{3}{2}$

The equation of tangent (with slope m) to the parabola $y^2 = 4ax$ is $\;$ $y = mx + \dfrac{a}{m}$

$\therefore$ $\;$ the required equation of tangent is

$y = \dfrac{3}{2} x + \dfrac{3 / 2}{3 / 2}$

i.e. $\;$ $y = \dfrac{3}{2}x + 1$

i.e. $\;$ $3x - 2y + 2 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equations of tangent and normal to the ellipse $x^2 + 4y^2 = 32$ at $\theta = \dfrac{\pi}{4}$

Equation of ellipse is $\;$ $x^2 + 4y^2 = 32$

i.e. $\;$ $\dfrac{x^2}{32} + \dfrac{y^2}{32 / 4} = 1$

i.e. $\;$ $\dfrac{x^2}{32} + \dfrac{y^2}{8} = 1$ $\;\;\; \cdots \; (1)$

The major axis of the ellipse given by equation $(1)$ is along the X axis.

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 32 \implies a = 4 \sqrt{2}$ $\;$ and $\;$ $b^2 = 8 \implies b = 2 \sqrt{2}$

$\theta = \dfrac{\pi}{4}$ represents the point $\left(a \cos \theta, b \sin \theta\right)$

i.e. the point $\left(4 \sqrt{2} \cos \left(\dfrac{\pi}{4}\right), 2 \sqrt{2} \sin \left(\dfrac{\pi}{4}\right)\right)$ $\;$ i.e. $\left(4,2\right)$

Equation of tangent to the ellipse at the point $\left(x_1, y_1\right)$ is $\;$ $\dfrac{xx_1}{a^2} + \dfrac{yy_1}{b^2} = 1$

Equation of normal to the ellipse at the point $\left(x_1, y_1\right)$ is $\;$ $\dfrac{a^2 x}{x_1} - \dfrac{b^2 y}{y_1} = a^2 - b^2$

Here $\left(x_1, y_1\right) = \left(4,2\right)$

$\therefore$ $\;$ The required equation of tangent to the ellipse is

$\dfrac{4x}{32} + \dfrac{2y}{8} = 1$

i.e. $\;$ $\dfrac{x}{8} + \dfrac{y}{4} = 1$

i.e. $\;$ $x + 2y - 8 = 0$

The required equation of normal to the ellipse is

$\dfrac{32x}{4} - \dfrac{8y}{2} = 32 - 8$

i.e. $\;$ $8x - 4y = 24$

i.e. $\;$ $2x - y - 6 = 0$

ALTERNATELY

Equation of tangent to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ in parametric form is $\;$ $\dfrac{x \cos \theta}{a} + \dfrac{y \sin \theta}{b} = 1$

Equation of normal to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ in parametric form is $\;$ $\dfrac{ax}{\cos \theta} - \dfrac{b y}{\sin \theta} = a^2 - b^2$

$\therefore$ $\;$ Required equation of tangent is

$\dfrac{x \cos \left(\dfrac{\pi}{4}\right)}{4 \sqrt{2}} + \dfrac{y \sin \left(\dfrac{\pi}{4}\right)}{2 \sqrt{2}} = 1$

i.e. $\;$ $\dfrac{x}{8} + \dfrac{y}{4} = 1$

i.e. $\;$ $x + 2y - 8 = 0$

Required equation of normal is

$\dfrac{4 \sqrt{2} x}{\cos \left(\dfrac{\pi}{4}\right)} - \dfrac{2 \sqrt{2} y}{\sin \left(\dfrac{\pi}{4}\right)} = 32 - 8$

i.e. $\;$ $8x - 4y = 24$

i.e. $\;$ $2x - y - 6 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equations of tangent and normal to the parabola $y^2 = 8x$ at $t = \dfrac{1}{2}$.

Given: Equation of parabola is $\;$ $y^2 = 8x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 8 \implies a = 2$

Equation of tangent to a parabola at a point $t$ is $\;$ $yt = x+ at^2$

Equation of normal to a parabola at a point $t$ is $\;$ $y + tx = 2at + at^3$

Given: $\;$ $t = \dfrac{1}{2}$

$\therefore$ $\;$ Required equation of tangent is:

$y \times \dfrac{1}{2} = x + 2 \times \dfrac{1}{4}$

i.e. $\;$ $2x - y + 1 = 0$

Required equation of normal is:

$y + \dfrac{x}{2} = 2 \times 2 \times \dfrac{1}{2} + 2 \times \dfrac{1}{8}$

i.e. $\;$ $y + \dfrac{x}{2} = \dfrac{9}{4}$

i.e. $\;$ $2x + 4y - 9 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equations of tangent and normal to the hyperbola $9x^2 - 5y^2 = 31$ at $\left(2,-1\right)$.

Equation of given hyperbola: $\;$ $9x^2 - 5y^2 = 31$

i.e. $\;$ $\dfrac{x^2}{31 / 9} - \dfrac{y^2}{31 / 5} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis of the hyperbola is along the X axis.

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = \dfrac{31}{9}$ $\;$ and $\;$ $b^2 = \dfrac{31}{5}$

Equation of tangent to the hyperbola at the point $\left(x_1, y_1\right)$ is

$\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1$

Given: $\left(x_1, y_1\right) = \left(2, -1\right)$

$\therefore$ $\;$ Required equation of tangent is

$\dfrac{2x}{31 / 9} + \dfrac{y}{31 / 5} = 1$

i.e. $\;$ $18x + 5y = 31$

Equation of normal to the hyperbola at the point $\left(x_1, y_1\right)$ is

$\dfrac{a^2 x}{x_1} + \dfrac{b^2 y}{y_1} = a^2 + b^2$

$\therefore$ $\;$ Required equation of normal is

$\dfrac{31 x}{9 \times 2} + \dfrac{31 y}{5 \times \left(-1\right)} = \dfrac{31}{9} + \dfrac{31}{5}$

i.e. $\;$ $\dfrac{x}{18} - \dfrac{y}{5} = \dfrac{14}{45}$

i.e. $\;$ $5x - 18 y - 28 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equations of tangent and normal to the parabola $x^2 + 2x - 4y + 4 = 0$ at $\left(0,1\right)$.

Equation of given parabola is: $\;$ $x^2 + 2x - 4y + 4 = 0$

i.e. $\;$ $x^2 + 2x = 4y - 4$

i.e. $\;$ $x^2 + 2x +1 = 4y - 4 + 1$

i.e. $\;$ $\left(x + 1\right)^2 = 4y - 3$

i.e. $\;$ $\left(x + 1\right)^2 = 4 \left(y - \dfrac{3}{4}\right)$ $\;\;\; \cdots \; (1)$

Let $\;$ $X = x + 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - \dfrac{3}{4}$ $\;\;\; \cdots \; (2b)$

Then, we have from equations $(1)$, $(2a)$ and $(2b)$,

$X^2 = 4Y$ $\;\;\; \cdots \; (3)$

Comparing equation $(3)$ with the standard equation of parabola $\;$ $X^2 = 4a Y$ $\;$ gives $\;$ $a = 1$

Given: Point $\left(x_1, y_1\right) = \left(0, 1\right)$

$\therefore$ $\;$ We have from equations $(2a)$ and $(2b)$,

when $x_1 = 0 \implies X_1 = 0 + 1 = 1$

and when $y_1 = 1 \implies Y_1 = 1 - \dfrac{3}{4} = \dfrac{1}{4}$

$\therefore$ $\;$ $\left(X_1, Y_1\right) = \left(1, \dfrac{1}{4}\right)$

Equation of tangent at $\left(X_1, Y_1\right)$ to the parabola given by equation $(3)$ is

$XX_1 = 2a \left(Y + Y_1\right)$

$\therefore$ $\;$ Equation of tangent to the parabola given by equation $(3)$ at the point $\left(X_1, Y_1\right)$ is

$1 \times X = 2 \times 1 \times \left(Y + \dfrac{1}{4}\right)$

i.e. $\;$ $X = 2Y + \dfrac{1}{2}$ $\;\;\; \cdots \; (4)$

Equation of normal at $\left(X_1, Y_1\right)$ to the parabola given by equation $(3)$ is

$YX_1 + 2a X = X_1 Y_1 + 2a X_1$

i.e. $\;$ $1 \times Y + 2 \times 1 \times X = 1 \times \dfrac{1}{4} + 2 \times 1 \times 1$

i.e. $\;$ $Y + 2X = \dfrac{9}{4}$ $\;\;\; \cdots \; (5)$

Substituting for $X$ and $Y$ from equations $(2a)$ and $(2b)$ in equation $(4)$, the equation of tangent to parabola given by equation $(1)$ is

$x +1 = 2 \left(y - \dfrac{3}{4}\right) + \dfrac{1}{2}$

i.e. $\;$ $x = 2y + \dfrac{1}{2} - \dfrac{3}{2} - 1$

i.e. $\;$ $x - 2y + 2 = 0$

Substituting for $X$ and $Y$ from equations $(2a)$ and $(2b)$ in equation $(5)$, the equation of normal to the parabola given by equation $(1)$ is

$y - \dfrac{3}{4} + 2 \times \left(x + 1\right) = \dfrac{9}{4}$

i.e. $\;$ $2x + y - \dfrac{3}{4} + 2 - \dfrac{9}{4} = 0$

i.e. $\;$ $2x + y - 1 = 0$

Application of Derivatives: Tangents and Normals

For the curve $y=x^2+3x+4$, find all the points at which the tangent passes through the origin.

$y=x^2+3x+4$ $\;\; \cdots$ (1)

Differentiating equation (1) w.r.t x gives

$\dfrac{dy}{dx}=2x+3$

The required tangent passes through the origin $\left(0,0\right)$

$\therefore$ $\;$ Equation of tangent is

$y-0 = \left(2x+3\right) \left(x-0\right)$

i.e. $y=2x^2 + 3x$ $\;\; \cdots$ (2)

Solving equations (1) and (2) simultaneously gives

$2x^2 + 3x = x^2 + 3x + 4$

i.e. $x^2 = 4$ $\implies$ $x = \pm 2$

Substituting the value of x in equation (1) gives

when $x=+2$, $y= \left(2\right)^2 + 3 \times 2 + 4 = 14$

when $x=-2$, $y = \left(-2\right)^2 + 3 \times \left(-2\right) + 4 = 2$

$\therefore$ $\;$ The required points are $\left(2,14\right)$ and $\left(-2,2\right)$

Application of Derivatives: Tangents and Normals

Find the measure of angle between the curves $x^2 - y^2 =3$ and $x^2 + y^2 -4x +3 = 0$

$x^2 -y^2 =3$ $\;\; \cdots$ (1)

$x^2 + y^2 -4x + 3 = 0$ $\;\; \cdots$ (2)

Adding equations (1) and (2) gives

$2x^2 -4x=0$

i.e. $x \left(x-2\right)=0$ $\implies$ $x=0$ or $x =2$

$\therefore$ $\;$ From equation (1),

when $x=0$, $y^2 = -3$

$\therefore$ $\;$ $x=0$ is not a valid answer.

when $x=2$, $y^2 = 1$ $\implies$ $y = \pm 1$

$\therefore$ $\;$ The two curves intersect at the points $\left(2,1\right)$ and $\left(2,-1\right)$

Differentiating equation (1) w.r.t x gives

$2x -2 y \dfrac{dy}{dx}=0$

i.e. $\dfrac{dy}{dx}=\dfrac{x}{y}$

$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(2,1\right)} = 2 = m_1 \left(\text{say}\right)$

and $\dfrac{dy}{dx}\bigg|_{\left(2,-1\right)}=-2 = m_3 \left(\text{say}\right)$

Differentiating equation (2) w.r.t x gives

$2x + 2y \dfrac{dy}{dx}-4=0$

i.e. $\dfrac{dy}{dx}= \dfrac{2-x}{y}$

$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(2,1\right)} = 0 = m_2 \left(\text{say}\right)$

and $\dfrac{dy}{dx}\bigg|_{\left(2,-1\right)}=0 = m_4 \left(\text{say}\right)$

Let $\alpha$ be the angle between the two tangents at the point $\left(2,1\right)$

Then $\tan \alpha = \left|\dfrac{m_1 - m_2}{1+m_1 m_2}\right|= \left|\dfrac{2-0}{1+2\times0}\right|=2$

$\implies$ $\alpha = \tan^{-1}\left(2\right)$

Let $\beta$ be the angle between the two tangents at the point $\left(2,-1\right)$

Then $\tan \beta = \left|\dfrac{m_3 - m_4}{1+m_3 m_4}\right|= \left|\dfrac{-2-0}{1+\left(-2\right)\times0}\right|=2$

$\implies$ $\beta = \tan^{-1}\left(2\right)$

Now, angle between the tangents of curves = angle between the curves

$\therefore$ Angle between the two curves $=\tan^{-1}\left(2\right)$

Application of Derivatives: Tangents and Normals

Show that the normal at any point $\theta$ to the curve $x=a \cos \theta + a \theta \sin \theta$, $y = a \sin \theta - a \theta \cos \theta$ is at a constant distance from the origin.

$x = a \cos \theta + a \theta \sin \theta$

$\therefore$ $\;$ $\dfrac{dx}{d\theta}=-a\sin \theta + a \sin \theta + a \theta \cos \theta$

i.e. $\dfrac{dx}{d\theta} = a \theta \cos \theta$ $\;\; \cdots$ (1)

$y = a \sin \theta - a \theta \cos \theta$

$\therefore$ $\;$ $\dfrac{dy}{d\theta} = a \cos \theta -a \cos \theta + a \theta \sin \theta$

i.e. $\dfrac{dy}{d\theta} = a \theta \sin \theta$ $\;\; \cdots$ (2)

$\therefore$ $\;$ $\dfrac{dy}{dx}= \dfrac{dy /d\theta}{dx / d \theta} = \dfrac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta$ $\;\;$ [from equations (1) and (2)]

$\therefore$ $\;$ Slope of normal $= \dfrac{-1}{dy / dx}=-\cot \theta$

$\therefore$ $\;$ Equation of normal is

$y - \left(a \sin \theta - a \theta \cos \theta\right) = -\cot \theta \left(x-a\cos \theta -a \theta \sin \theta\right)$

i.e. $\sin \theta\left(y-a\sin \theta + a\theta cos \theta\right) = -\cos \theta \left(x-a\cos\theta - a \theta \sin \theta\right)$

i.e. $y \sin \theta -a \sin^2 \theta + a \theta \sin \theta \cos \theta = -x \cos \theta + a \cos^2 \theta + a \theta \sin \theta \cos \theta$

i.e. $x \cos \theta + y \sin \theta=a \left(\sin^2 \theta + \cos^2 \theta\right)$

i.e. $x \cos \theta + y \sin \theta - a = 0$

$\therefore$ $\;$ Distance of the normal from the origin $= \left|\dfrac{-a}{\sqrt{\cos^2 \theta + \sin^2 \theta}}\right|=a$

$\therefore$ $\;$ The normal is at a constant distance from the origin.

Note: Distance of the line $Ax+By+C=0$ from the point $\left(x_0,y_0\right)$ is $\left|\dfrac{Ax_0 + By_0 + C}{\sqrt{A^2 + B^2}}\right|$

Application of Derivatives: Tangents and Normals

Show that the curves $xy=a^2$ and $x^2+y^2=2a^2$ touch each other.

The two curves are

$xy=a^2$ $\implies$ $x = \dfrac{a^2}{y}$ $\;\; \cdots$ (1)

$x^2 + y^2 = 2a^2$ $\implies$ $\left(\dfrac{a^2}{y}\right)^2 + y^2 = 2a^2$ [in view of equation (1)]

i.e. $a^4 + y^4 = 2a^2 y^2$

i.e. $y^4 - 2a^2 y^2 +a^4 =0$

i.e. $\left(y^2-a^2\right)^2 = 0$

i.e. $y^2 -a^2 =0$ $\implies$ $y^2 = a^2$ $\implies$ $y = \pm a$

$\therefore$ $\;$ From equation (1),

when $y = a$, $x=a$ and

when $y = -a$, $x=-a$

$\therefore$ $\;$ The given curves intersect each other at the points $\left(a,a\right)$ and $\left(-a,-a\right)$

Differentiating $xy=a^2$ w.r.t x gives

$x \dfrac{dy}{dx}+y=0$

i.e. $\dfrac{dy}{dx}=\dfrac{-y}{x}$

$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(a,a\right)}= \dfrac{-a}{a}=-1=m_1 \left(\text{say}\right)$ $\;\; \cdots$ (2)

Differentiating $x^2 + y^2 = 2a^2$ w.r.t x gives

$2x+2y\dfrac{dy}{dx}=0$

i.e. $\dfrac{dy}{dx}=\dfrac{-x}{y}$

$\therefore$ $\;$ $\dfrac{dy}{dx}\bigg|_{\left(a,a\right)}= \dfrac{-a}{a}=-1 = m_2 \left(\text{say}\right)$ $\;\; \cdots$ (3)

Let $\phi$ be the angle between the two tangents.

Then $\tan \phi = \left|\dfrac{m_1 - m_2}{1+m_1 m_2}\right| = \left|\dfrac{-1+1}{1+\left(-1\right)\left(-1\right)}\right|=0$

$\implies$ $\phi = 0$

i.e. the two tangents touch each other.

$\implies$ The two curves touch each other.

Application of Derivatives: Tangents and Normals

Show that the curves $4x=y^2$ and $4xy=k$ cut at right angle if $k^2=512$

Since the curves cut at right angle $\implies$ their tangents are perpendicular to each other

Consider $4x=y^2$

Differentiating w.r.t x gives

$4= 2y \dfrac{dy}{dx}$

$\implies$ $\dfrac{dy}{dx}=\dfrac{2}{y}$

Consider $4xy=k$

Differentiating w.r.t x gives

$4\left(x\dfrac{dy}{dx}+y\right)=0$

$\implies$ $\dfrac{dy}{dx}=-\dfrac{y}{x}$

Since the tangents are perpendicular to each other

$\implies$ $\dfrac{2}{y}\times \left(\dfrac{-y}{x}\right)=-1$ $\implies$ $x=2$

Substituting the value of x in the equation of the curve $4xy=k$ gives

$8y=k$ $\implies$ $y = \dfrac{k}{8}$

Substituting the values of x and y in the equation of the curve $4x=y^2$ gives

$4 \times 2 = \left(\dfrac{k}{8}\right)^2$

$\implies$ $k^2=8\times 64 = 512$

$\therefore$ $\;$ The two curves cut at right angle if $k^2 = 512$

Application of Derivatives: Tangents and Normals

Find the equation of tangent to the curve $x^2-2y^2=8$ which are perpendicular to the line $x-y+29=0$

Slope of the line $x-y+29=0$ is $m=1$

Since the required tangent is perpendicular to the given line

slope of tangent $=-\dfrac{1}{m}=-1$ $\;\; \cdots$ (1)

Differentiating the equation of the curve $x^2-2y^2=8$ with respect to x gives

$2x-4y \dfrac{dy}{dx}=0$

i.e. $2y \dfrac{dy}{dx}=x$

$\implies$ $\dfrac{dy}{dx}=\dfrac{x}{2y}$ $\;\; \cdots$ (2)

$\therefore$ $\;$ We have from equations (1) and (2)

$\dfrac{x}{2y}=-1$

$\implies$ $x=-2y$ $\;\; \cdots$ (3)

In view of equation (3), the equation of the curve becomes

$\left(-2y\right)^2-2y^2=8$

i.e. $2y^2=8$ $\implies$ $y^2=4$ $\implies$ $y=\pm 2$

Substituting the value of y in equation (3) gives

$x=-2y=\mp 4$

$\therefore$ $\;$ The tangents to the given curve are located at the points $\left(-4,2\right)$, $\left(4,-2\right)$

$\therefore$ $\;$ Equation of tangent at $\left(-4,2\right)$ is

$y-2=-1\left(x+4\right)$ $\implies$ $x+y+2=0$

$\therefore$ $\;$ Equation of tangent at $\left(4,-2\right)$ is

$y+2=-1\left(x-4\right)$ $\implies$ $x+y-2=0$

$\therefore$ $\;$ The two tangents to the given curve are $x+y+2=0$ and $x+y-2=0$

Application of Derivatives: Tangents and Normals

Find the equation of tangent to $x=\cos \theta$, $y=\sin \theta$ $\;\;$ $\theta \in \left[0,2\pi\right)$ at $\theta=\dfrac{\pi}{4}$

$\dfrac{dx}{d\theta}=-\sin \theta$

$\dfrac{dy}{d\theta}=\cos \theta$

$\therefore$ $\;\;$ $\dfrac{dy}{dx}=\dfrac{dy / d\theta}{dx / d\theta} = -\dfrac{\cos \theta}{\sin \theta}=-\cot \theta$

$\therefore$ $\;\;$ $\dfrac{dy}{dx}\bigg|_{\theta=\frac{\pi}{4}} = -\cot \left(\dfrac{\pi}{4}\right)=-1$

When $\theta = \dfrac{\pi}{4}$,

$x = \cos \left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$

$y = \sin \left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$

$\therefore$ $\;\;$ Equation of tangent at $\theta=\dfrac{\pi}{4}$ is

$y-\dfrac{1}{\sqrt{2}}=-1 \left(x-\dfrac{1}{\sqrt{2}}\right)$

i.e. $x\sqrt{2} + y \sqrt{2}=2$

Application of Derivatives: Tangents and Normals

The curve $y=ax^3+bx^2+cx+5$ touches the X axis at the point $\left(-2,0\right)$ and cuts the Y axis at a point where its gradient is 3. Find a, b and c.

When the curve $y=ax^3+bx^2+cx+5$ cuts the Y axis, the X coordinate of the point is 0.

$\implies$ $y=5$

$\therefore$ $\;\;$ At the point $\left(0,5\right)$, the gradient of the curve is 3

Differentiating the curve $y=ax^3+bx^2+cx+5$ with respect to x gives

$\dfrac{dy}{dx}=3ax^2+2bx+c \;\; \cdots$ (1)

$\therefore$ $\;\;$ $\dfrac{dy}{dx}\bigg|_{\left(0,5\right)}=c$

The gradient of the curve at $\left(0,5\right)$ is 3

$\implies$ $c=3 \;\; \cdots$ (2)

The curve $y=ax^3+bx^2+cx+5$ touches the X axis at the point $\left(-2,0\right)$

$\implies$ $0=\left(-2\right)^3 a + \left(-2\right)^2 b -2c+5$

i.e. $8a-4b+2c = 5$

Substituting the value of c gives

$8a-4b=-1 \;\; \cdots$ (3)

Since the curve touches the X axis at $\left(-2,0\right)$

$\implies$ X axis is tangent to the curve at $\left(-2,0\right)$

$\implies$ $\dfrac{dy}{dx}\bigg|_{\left(-2,0\right)}=0$

$\therefore$ $\;\;$ We have from equations (1) and (2)

$3 \times \left(-2\right)^2 a + 2 \times \left(-2\right)b + 3 = 0$

i.e. $12a-4b=-3 \;\; \cdots$ (4)

Solving equations (3) and (4) simultaneously gives

$4a = -2$ $\implies$ $a = -\dfrac{1}{2}$

Substituting the value of a in equation (3) gives

$8 \times \left(\dfrac{-1}{2}\right)-4b=-1$

i.e. $4b = -3$ $\implies$ $b = -\dfrac{3}{4}$

$\therefore$ $\;\;$ $a=-\dfrac{1}{2}$, $b = -\dfrac{3}{4}$, $c=3$