Area Ques 68
Question
- The tangent to the parabola $y^{2}=4 x$ at the point where it intersects the circle $x^{2}+y^{2}=5$ in the first quadrant, passes through the point
(a) $(\frac{1}{4}, \frac{3}{4})$
(b) $(\frac{3}{4}, \frac{7}{4})$
(c) $(-\frac{1}{3}, \frac{4}{3})$
(d) $(-\frac{1}{4}, \frac{1}{2})$
Show Answer
Answer:
Correct Answer: 68.(b)
Solution:
Formula:
- Given equations of the parabola $y^{2}=4 x$ and circle $\quad$ …….(i)
$ x^{2}+y^{2}=5 $ $\quad$ …….(ii)
So, for point of intersection of curves (i) and (ii), put $y^{2}=4 x$ in Eq. (ii), we get
$ \begin{array}{lc} & \\ \Rightarrow & x^{2}+4 x-5=0 \\ \Rightarrow & x^2 + 5x -x -5=0 \\ \Rightarrow & (x-1)(x+5)=0 \\ \Rightarrow & x=1,-5 \end{array} $
For first quadrant $x=1$, so $y=2$.
Now, equation of tangent of parabola (i) at point $(1,2)$ is $T=0$
$\Rightarrow \quad 2 y=2(x+1)$
$\Rightarrow \quad x-y+1=0$
The point $(\frac{3}{4}, \frac{7}{4})$ satisfies, the equation of line
$ x-y+1=0 $
BETA
