Let P be a point on the ellipse \fracx^29+\fracy^24=1 . Let the line passing through P and parallel to y -axis meet the circle x^2+y^2=9 at point Q such that P and Q are on the same side of the x -axis. Then, the eccentricity of the locus of the point R on P Q such that P R: R Q=4: 3 as P moves on the ellipse, is :

Ellipse PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q39 with Solution

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Correct answer: C

Coordinates of $P$ and $Q$

Let $P=(x,y_1)$ on the ellipse $\frac{x^2}{9}+\frac{y_1^2}{4}=1.$

The line through $P$ parallel to the $y$ -axis is $x=\text{constant}=x$ . It meets the circle $x^2+y^2=9$ at point $Q=(x,y_2)$ .

Since $P$ and $Q$ are on the same side of the $x$ -axis, $y_1$ and $y_2$ have the same sign. Hence we take $y_1=\pm 2\sqrt{1-\frac{x^2}{9}}, \qquad y_2=\pm \sqrt{9-x^2}$ with the same sign.

Now, $y_1^2=4\left(1-\frac{x^2}{9}\right)=4-\frac{4x^2}{9},$ so $y_1=\pm \frac{2}{3}\sqrt{9-x^2}.$ Since $y_1,y_2$ have the same sign, $y_1=\frac{2}{3}y_2 \quad \Rightarrow \quad y_2=\frac{3}{2}y_1.$

Coordinates of $R$

Point $R$ divides $PQ$ internally in the ratio $PR:RQ=4:3.$

Using section formula, $R=\left(\frac{4x+3x}{7},\frac{4y_2+3y_1}{7}\right)?$ Be careful: if $PR:RQ=m:n$ , then $R=\left(\frac{nx_P+mx_Q}{m+n},\frac{ny_P+my_Q}{m+n}\right).$ Here $m=4,n=3$ , so $R=\left(\frac{3x_P+4x_Q}{7},\frac{3y_1+4y_2}{7}\right).$ Since $x_P=x_Q=x$ , $x_R=x.$ Also,

Using $y_2=\frac{3}{2}y_1$ ,

=\frac{3y_1+6y_1}{7} =\frac{9}{7}y_1.$$ So, $$y_1=\frac{7}{9}y_R.$$ --- 3. **Equation of the locus of** $R$ Since $P$ lies on the ellipse, $$\frac{x^2}{9}+\frac{y_1^2}{4}=1.$$ Substitute $y_1=\frac{7}{9}y_R$ and $x=x_R$: $$\frac{x_R^2}{9}+\frac{1}{4}\left(\frac{7y_R}{9}\right)^2=1.$$ Thus, $$\frac{x_R^2}{9}+\frac{49y_R^2}{324}=1.$$ Rewrite as $$\frac{x_R^2}{9}+\frac{y_R^2}{\frac{324}{49}}=1.$$ So the locus is an ellipse with semi-axes $$a=3, \qquad b=\frac{18}{7}.$$ Since $3>\frac{18}{7}$, major axis is along $x$-axis. --- 4. **Eccentricity** For ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a>b,$$ eccentricity is $$e=\sqrt{1-\frac{b^2}{a^2}}.$$ Here, $$a^2=9, \qquad b^2=\frac{324}{49}.$$ Therefore, $$e=\sqrt{1-\frac{324/49}{9}} =\sqrt{1-\frac{36}{49}} =\sqrt{\frac{13}{49}} =\frac{\sqrt{13}}{7}.$$ --- 5. **Option check** The eccentricity is $$\boxed{\frac{\sqrt{13}}{7}}$$ which matches **Option C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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Correct answer: C

Coordinates of $P$ and $Q$

Let $P=(x,y_1)$ on the ellipse $\frac{x^2}{9}+\frac{y_1^2}{4}=1.$

The line through $P$ parallel to the $y$ -axis is $x=\text{constant}=x$ . It meets the circle $x^2+y^2=9$ at point $Q=(x,y_2)$ .

Since $P$ and $Q$ are on the same side of the $x$ -axis, $y_1$ and $y_2$ have the same sign. Hence we take $y_1=\pm 2\sqrt{1-\frac{x^2}{9}}, \qquad y_2=\pm \sqrt{9-x^2}$ with the same sign.

Now, $y_1^2=4\left(1-\frac{x^2}{9}\right)=4-\frac{4x^2}{9},$ so $y_1=\pm \frac{2}{3}\sqrt{9-x^2}.$ Since $y_1,y_2$ have the same sign, $y_1=\frac{2}{3}y_2 \quad \Rightarrow \quad y_2=\frac{3}{2}y_1.$

Coordinates of $R$

Point $R$ divides $PQ$ internally in the ratio $PR:RQ=4:3.$

Using $y_2=\frac{3}{2}y_1$ ,

=\frac{3y_1+6y_1}{7} =\frac{9}{7}y_1.$$ So, $$y_1=\frac{7}{9}y_R.$$ --- 3. **Equation of the locus of** $R$ Since $P$ lies on the ellipse, $$\frac{x^2}{9}+\frac{y_1^2}{4}=1.$$ Substitute $y_1=\frac{7}{9}y_R$ and $x=x_R$: $$\frac{x_R^2}{9}+\frac{1}{4}\left(\frac{7y_R}{9}\right)^2=1.$$ Thus, $$\frac{x_R^2}{9}+\frac{49y_R^2}{324}=1.$$ Rewrite as $$\frac{x_R^2}{9}+\frac{y_R^2}{\frac{324}{49}}=1.$$ So the locus is an ellipse with semi-axes $$a=3, \qquad b=\frac{18}{7}.$$ Since $3>\frac{18}{7}$, major axis is along $x$-axis. --- 4. **Eccentricity** For ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a>b,$$ eccentricity is $$e=\sqrt{1-\frac{b^2}{a^2}}.$$ Here, $$a^2=9, \qquad b^2=\frac{324}{49}.$$ Therefore, $$e=\sqrt{1-\frac{324/49}{9}} =\sqrt{1-\frac{36}{49}} =\sqrt{\frac{13}{49}} =\frac{\sqrt{13}}{7}.$$ --- 5. **Option check** The eccentricity is $$\boxed{\frac{\sqrt{13}}{7}}$$ which matches **Option C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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