Circle, Parabola, Ellipse & Hyperbola — Mixed
DAILY_CHALLENGE
Grade None
Question:
Match each entry in List-I to the correct entry in List-II and choose the correct option.
**List-I**
(P) The circle with centre $(1,2)$ and touching the straight line $3x+4y=1$ passes through
(Q) The common tangent to the circle $x^2+y^2=2$ and the parabola $y^2=8x$ with positive slope passes through
(R) Let $M$ be the end point of the latus rectum of the ellipse $3x^2+4y^2=48$ such that $M$ lies in the first quadrant. Then the normal to the ellipse drawn at $M$ passes through
(S) Let $H$ be the hyperbola whose centre is at the origin, one of the foci is at $(5,0)$, and one directrix is $5x+16=0$. Then $H$ passes through
**List-II**
(1) the point $(1,1)$
(2) the point $(7,9)$
(3) the point $(3,2)$
(4) the point $(2,5)$
(5) the point $(8, 3\sqrt{3})$
$(P)\to(3),\ (Q)\to(4),\ (R)\to(1),\ (S)\to(2)$
$(P)\to(3),\ (Q)\to(2),\ (R)\to(1),\ (S)\to(5)$
$(P)\to(3),\ (Q)\to(2),\ (R)\to(4),\ (S)\to(5)$
$(P)\to(4),\ (Q)\to(1),\ (R)\to(2),\ (S)\to(3)$
Step-by-Step Solution
Key Concept: Each part tests a distinct conic property: (P) radius = distance from centre to tangent; (Q) double-tangency condition; (R) standard normal equation $\frac{a^2x}{x_0}-\frac{b^2y}{y_0}=a^2-b^2$; (S) focus-directrix relation $ae \cdot \frac{a}{e} = a^2$.
**Step 1: P: Circle with centre $(1,2)$ touching $3x+4y=1$**
Radius $= \dfrac{|3(1)+4(2)-1|}{\sqrt{9+16}} = \dfrac{10}{5} = 2$. Circle: $(x-1)^2+(y-2)^2=4$. Check $(3,2)$: $4+0=4$. ✓ $P\to(3)$.
**Step 2: Q: Common tangent (positive slope) to $x^2+y^2=2$ and $y^2=8x$**
Tangent to $y^2=8x$ with slope $m$: $y=mx+\tfrac{2}{m}$. Tangency with $x^2+y^2=2$: $\dfrac{|2/m|}{\sqrt{1+m^2}}=\sqrt{2} \Rightarrow m^4+m^2-2=0 \Rightarrow m=1$. Tangent: $y=x+2$. Check $(7,9)$: $9=7+2$. ✓ $Q\to(2)$.
**Step 3: R: Normal to ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$ at latus rectum endpoint**
$c=2$. Latus rectum endpoint in Q1: $(2, 3)$. Normal: $\dfrac{16x}{2}-\dfrac{12y}{3}=16-12 \Rightarrow 8x-4y=4 \Rightarrow 2x-y=1$. Check $(1,1)$: $2-1=1$. ✓ $R\to(1)$.
**Step 4: S: Hyperbola with focus $(5,0)$, directrix $x=-16/5$**
$c=5$, $a/e=16/5$, $ae=5 \Rightarrow a^2=16, e=5/4, b^2=9$. Hyperbola: $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$. Check $(8,3\sqrt{3})$: $\tfrac{64}{16}-\tfrac{27}{9}=4-3=1$. ✓ $S\to(5)$.
Correct Answer: B