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