Question

If the straight line $y=4x+c$ touches the ellipse $\frac{x^2}{4}+y^2=1$ then c is equal to

• A. 0
• B. $\pm \sqrt{65}$
• C. $\pm \sqrt{62}$
• D. $\pm \sqrt{2}$
• E. $\pm 13$

Answer 1

Answer: (B)

We have,
$y=4x+c...(i)$
and $\frac{x^2}{4}+y^2=1...(ii)$
Put value of $y$ from Eqs. (i) into (ii), we get
$\frac{x^2}{4}+(4x+c{)}^2=1$
$\Rightarrow x^2+4(4x+c{)}^2=4$
$\Rightarrow x^2+4\left(16x^2+8cx+c^2\right)=4$
$\Rightarrow x^2+64x^2+32cx+4c^2=4$
$\Rightarrow 65x^2+32cx+4\left(c^2-1\right)=0$
Since, given line is a tangent to the ellipse.
$\therefore$ Discriminant $=0$
$\Rightarrow (32c{)}^2-4\times 65\times 4\left(c^2-1\right)=0$
$\Rightarrow 1024c^2-1040\left(c^2-1\right)=0$
$\Rightarrow 1024c^2-1040c^2+1040=0$
$\Rightarrow 16c^2=1040$
$\Rightarrow c^2=65$
$\Rightarrow c=\pm \sqrt{65}$