Ellipse Ques 22
- The number of values of $c$ such that the straight line $y=4 x+c$ touches the curve $\frac{x^{2}}{4}+y^{2}=1$ is
(a) 0
(b) 2
(c) 1
(d) $\infty$
$(1998,2 M)$
Show Answer
Answer:
Correct Answer: 22.(b)
Solution:
- For ellipse, condition of tangency is $c^{2}=a^{2} m^{2}+b^{2}$
Given line is $y=4 x+c$ and curve $\frac{x^{2}}{4}+y^{2}=1$
$ \begin{array}{ll} \Rightarrow & c^{2}=4 \times 4^{2}+1=65 \\ \Rightarrow & c= \pm \sqrt{65} \end{array} $
So, there are two different values of $C$.
BETA
