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 :
Correct Option: , 3
Solution:
To find intersection point of $x^{2}+y^{2}=5$ and $y^{2}=4 x$,
substitute $y^{2}=4 x$ in $x^{2}+y^{2}=5$, we get
$x^{2}+4 x-5=0 \Rightarrow x^{2}+5 x-x-5=0$
$\Rightarrow x(x+5)-1(x+5)=0$
$\therefore x=1,-5$
Intersection point in $1^{\text {st }}$ quadrant be $(1,2)$.
Now, equation of tangent to $y^{2}=4 x$ at $(1,2)$ is
$y \times 2=2(x+1) \Rightarrow y=x+1$
$\Rightarrow x-y=0$ .....$\ldots(\mathrm{i})$
Hence, $\left(\frac{3}{4}, \frac{7}{4}\right)$ lies on (i)