Area Question 7
Question 7
- 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) $-\frac{1}{4}, \frac{1}{2}$
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Solution:
- Given equations of the parabola $y^{2}=4 x$ and circle
$$ x^{2}+y^{2}=5 $$
So, for point of intersection of curves (i) and (ii), put $y^{2}=4 x$ in Eq. (ii), we get
$$ \begin{array}{lc} & \ \Rightarrow & x^{2}+4 x-5=0 \ \Rightarrow & (x-1)(x+5)=0 \ \Rightarrow & x=1,-5 \end{array} $$
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
$$ x-y+1=0 $$
