Question

Find the equations of the two tangents that can be drawn from

(5, 2)

to the ellipse

2 x^{2} + 7 y^{2} = 14

.

Accepted Answer

Given that, the equation is $2 x^{2} + 7 y^{2} = 14$ .

Dividing with $14$ on both sides.

$\Rightarrow \frac{x^{2}}{7} + \frac{y^{2}}{2} = 1$

On comparing with the standard ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , we get

$a = \sqrt{7}$ and $b = \sqrt{2}$

Equation of the tangent to the ellipse is $y = m x + \sqrt{a^{2} m^{2} + b^{2}}$ .

It also passes through the point $(5, 2)$ .

$\Rightarrow 2 = 5 m + \sqrt{7 m^{2} + 2}$

$\Rightarrow (2 - 5 m)^{2} = (7 m^{2} + 2)$

$\Rightarrow 4 + 25 m^{2} - 20 m = 7 m^{2} + 2$

$\Rightarrow 18 m^{2} - 20 m + 2 = 0$

$\Rightarrow (9 m - 1) (m - 1) = 0$

$\Rightarrow m = 1$ or $m = \frac{1}{9}$ .

When $m = 1$ ,

Equation of the line with slope $m$ and passing through point $(x_{1}, y_{1})$ is $(y - y_{1}) = m (x - x_{1})$ .

$\Rightarrow (y - 2) = 1 (x - 5)$

$\Rightarrow x - y - 3 = 0$

When $m = \frac{1}{9}$ ,

Equation of tangent is $y - 2 = \frac{1}{9} (x - 5)$

$\Rightarrow 9 y - 18 = x - 5$

$\Rightarrow x - 9 y + 13 = 0$

$∴$ The equation of the tangents is $x - y - 3 = 0$ and $x - 9 y + 13 = 0$ .

EXERCISE

Tamil Nadu Board Mathematics Solutions for EXERCISE

Questions from EXERCISE with Hints & Solutions

Learn and Prepare for any exam you want !