Q

Solve it, - Co-ordinate geometry - JEE Main-9

The tangent to the parabola $y^{2}=4x$at the point where it intersects the circle $x^{2}+y^{2}=5$ in the first quadrant, passes through the point :

• Option 1)

$\left ( -\frac{1}{3},\frac{4}{3} \right )$

• Option 2)

$\left ( \frac{1}{4},\frac{3}{4} \right )$

• Option 3)

$\left ( \frac{3}{4},\frac{7}{4} \right )$

• Option 4)

$\left ( -\frac{1}{4},\frac{1}{2} \right )$

Views

Point of intersection of $y^{2}=4x$ and $x^{2}+y^{2}=5$ is $(1,2)$ and $(1,-2)$

Equation of tangent at (1, 2)

$2y=2(x+1)\Rightarrow y=x+1$

$y-x=1$    point $\left ( \frac{3}{4} , \frac{7}{4}\right )$ satisfies .

Option 1)

$\left ( -\frac{1}{3},\frac{4}{3} \right )$

Option 2)

$\left ( \frac{1}{4},\frac{3}{4} \right )$

Option 3)

$\left ( \frac{3}{4},\frac{7}{4} \right )$

Option 4)

$\left ( -\frac{1}{4},\frac{1}{2} \right )$

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