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. $\pm \sqrt{2}$

Answer 1

Answer: (B)

$\pm \sqrt{65}$

Answer 2

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