The tangent to the parabola y^2 = 4x at the p | vedclass

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(B) Given equations are $y^2 = 4x$ and $x^2 + y^2 = 5$.Substituting $y^2 = 4x$ into the circle equation: $x^2 + 4x = 5$.$x^2 + 4x - 5 = 0 \Rightarrow

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

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(B) Given equations are $y^2 = 4x$ and $x^2 + y^2 = 5$.Substituting $y^2 = 4x$ into the circle equation: $x^2 + 4x = 5$.$x^2 + 4x - 5 = 0 \Rightarrow (x + 5)(x - 1) = 0$.Since the intersection is in the first quadrant,$x = 1$.For $x = 1$,$y^2 = 4(1) = 4$,so $y = 2$ (as $y > 0$ in the first quadrant).The point of intersection is $P(1, 2)$.The equation of the tangent to $y^2 = 4x$ at $(x_1, y_1)$ is $yy_1 = 2(x + x_1)$.Substituting $(1, 2)$: $2y = 2(x + 1) \Rightarrow y = x + 1$.Checking the options,for $x = \frac{3}{4}$,$y = \frac{3}{4} + 1 = \frac{7}{4}$.Thus,the tangent passes through $\left( \frac{3}{4}, \frac{7}{4} \right)$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

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

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