Question

The number of values of $c$ such that the straight line $y=4x+c$ touches the curve $\frac{x^2}{4}+y^2=1$ is :

• A. $0$
• B. $1$
• C. $2$
• D. Infinite

Answer 1

The correct option is C $2$

Since the line $y=4x+c$ touches the ellipse $\frac{x^2}{4}+y^2=1$, we substitute $y$ as $4x+c$ into the ellipse.

We get $\frac{x^2}{4}+16x^2+8cx+c^2=1$

$\Rightarrow x^2+64x^2+32cx+4c^2-4=0$

The discriminant of this quadratic should be zero since we have only one value of $x$.

$\Rightarrow (32c{)}^2=4\times 65\times (4c^2-4)$

$\Rightarrow 1024c^2=16\times 65\times (c^2-1)$

$\Rightarrow 64c^2=65\times (c^2-1)$

$\Rightarrow 64c^2=65c^2-65$

$\Rightarrow c^2=65$

Thus, we have two values of $c$.