Question

The equation of the tangent to the ellipse $4{{x}^{2}}+3{{y}^{2}}=5$ which is parallel to the straight line $y=3x+7$ is
A. $2\sqrt{3}x-6\sqrt{3}y+\sqrt{155}=0$
B. $2\sqrt{3}x-2\sqrt{3}y-\sqrt{155}=0$
C. $6\sqrt{3}x-2\sqrt{3}y+\sqrt{155}=0$
D. $6\sqrt{3}x-2\sqrt{3}y-\sqrt{155}=0$

Answer 1

Hint: Firstly, find the slope (m) by comparing with the line equation. Then compare the tangent equation with the given equation and find the value of A and B. Now put the value of a, and the equation of ellipse of the tangent.

Complete step-by-step answer:
Equation of line is $y=mx+c$
When compared with $y=3x+7$
Slope $\Rightarrow m=3$
Equation of tangent with slope m for ellipse
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
So, for the given ellipse the equation of tangent
$y=3x\pm \sqrt{\dfrac{45}{4}+\dfrac{5}{3}}$
$4{{x}^{2}}+3{{y}^{2}}=5$
$\dfrac{4{{x}^{2}}}{5}+\dfrac{3{{y}^{2}}}{5}=1$
$\dfrac{{{x}^{2}}}{\left( \dfrac{5}{4} \right)}+\dfrac{{{y}^{2}}}{\left( \dfrac{5}{3} \right)}=1$ so, ${{a}^{2}}=\dfrac{5}{4}$ and ${{b}^{2}}=\dfrac{5}{3}$.
$y=3x\pm \sqrt{\dfrac{155}{12}}$
On solving, $6\sqrt{3}x-2\sqrt{3}y\pm \sqrt{155}=0$

Note: On solving this question one should know the concept of ellipse, tangent and slope. It is very important for solving. In this we have to solve both the equations separately and find the values by comparison with the original equation.