Matching List Set on Circles Parabolas Ellipses and Hyperbolas

To find the correct matches between List-I and List-II, we solve each part step-by-step:

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Part (P):

The equation of a circle with centre $(1, 2)$ touching the straight line $3x + 4y = 1$ (or $3x + 4y - 1 = 0$) can be found by determining its radius $r$.

The radius $r$ is the perpendicular distance from the centre $(1, 2)$ to the given line: $r = \left| \frac{3(1) + 4(2) - 1}{\sqrt{3^2 + 4^2}} \right| = \frac{10}{5} = 2$

Thus, the equation of the circle is: $(x - 1)^2 + (y - 2)^2 = 2^2 = 4$

Checking the points from List-II:

• For point $(3, 2)$: $(3 - 1)^2 + (2 - 2)^2 = 2^2 + 0 = 4$ This point satisfies the equation of the circle.

$\therefore (\text{P}) \rightarrow (3)$

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Part (Q):

The equation of the parabola is $y^2 = 8x$, where $a = 2$.

The equation of a line with slope $m$ tangent to the parabola $y^2 = 8x$ is: $y = mx + \frac{2}{m} \implies m^2 x - my + 2 = 0$

Since this line is also tangent to the circle $x^2 + y^2 = 2$, the perpendicular distance from the centre $(0, 0)$ to the line must equal the radius $\sqrt{2}$: $\frac{|2|}{\sqrt{m^4 + m^2}} = \sqrt{2} \implies m^4 + m^2 = 2$

Solving for $m$: $m^4 + m^2 - 2 = 0 \implies (m^2 + 2)(m^2 - 1) = 0$

Since $m$ is real and given to have a positive slope ($m > 0$), we get: $m^2 = 1 \implies m = 1$

The equation of the common tangent is: $y = x + 2$

Checking the points from List-II:

• For point $(7, 9)$: $9 = 7 + 2$ This point lies on the line.

$\therefore (\text{Q}) \rightarrow (2)$

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Part (R):

The equation of the ellipse is $3x^2 + 4y^2 = 48$, which can be written in standard form as: $\frac{x^2}{16} + \frac{y^2}{12} = 1$

Here, $a^2 = 16 \implies a = 4$ and $b^2 = 12 \implies b = 2\sqrt{3}$.

The eccentricity $e$ is given by: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{12}{16}} = \frac{1}{2}$

The point $M$ is the end point of the latus rectum in the first quadrant: $x_1 = ae = 4 \times \frac{1}{2} = 2$ $y_1 = \frac{b^2}{a} = \frac{12}{4} = 3$ So, $M = (2, 3)$.

The equation of the normal to the ellipse at $(x_1, y_1) = (2, 3)$ is: $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$ $\frac{16x}{2} - \frac{12y}{3} = 16 - 12 \implies 8x - 4y = 4 \implies 2x - y = 1$

Checking the points from List-II:

• For point $(1, 1)$: $2(1) - 1 = 1$ This point satisfies the normal equation.

$\therefore (\text{R}) \rightarrow (1)$

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Part (S):

For the hyperbola $H$ centered at the origin:

• One focus is at $(5, 0) \implies ae = 5$
• One directrix is $5x + 16 = 0 \implies x = -\frac{16}{5} \implies \frac{a}{e} = \frac{16}{5}$

Multiplying the two relations: $a^2 = (ae) \left( \frac{a}{e} \right) = 5 \times \frac{16}{5} = 16 \implies a = 4$

Since $ae = 5$, we have $e = \frac{5}{4}$.

Now, finding $b^2$: $b^2 = a^2(e^2 - 1) = 16 \left( \frac{25}{16} - 1 \right) = 9$

The equation of the hyperbola is: $\frac{x^2}{16} - \frac{y^2}{9} = 1$

Checking the points from List-II:

• For point $(8, 3\sqrt{3})$: $\frac{8^2}{16} - \frac{(3\sqrt{3})^2}{9} = \frac{64}{16} - \frac{27}{9} = 4 - 3 = 1$ This point lies on the hyperbola.

$\therefore (\text{S}) \rightarrow (5)$

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Conclusion:

Matching List-I to List-II gives: $(\text{P}) \rightarrow (3),\ (\text{Q}) \rightarrow (2),\ (\text{R}) \rightarrow (1),\ (\text{S}) \rightarrow (5)$

This corresponds to Option B.