A vertical line passing through the point (h,0) intersects the ellipse x^2 \over 4 + y^2 \over 3 = 1 at the points P and Q . Let the tangents to the ellipse at P and Q meet at the point R . If \Delta ( h ) = area of the triangle PQR , \Delta _1 = \mathop \max \limits_1/2 \le h \le 1 \Delta ( h ) and \Delta _2 = \mathop \min \limits_1/2 \le h \le 1 \Delta ( h ) , then 8 \over \sqrt 5 \Delta _1 - 8\Delta _2 =

Ellipse PYQ — JEE Advanced 2013 · Shift 1 · Q30 with Solution

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Correct answer: 9

Equation of the ellipse and points of intersection

The ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1.$

A vertical line through $(h,0)$ is $x=h.$

It intersects the ellipse at points $P$ and $Q$ . Substituting $x=h$ into the ellipse: $\frac{h^2}{4}+\frac{y^2}{3}=1$ $\frac{y^2}{3}=1-\frac{h^2}{4}$ $y^2=3\left(1-\frac{h^2}{4}\right)=\frac{3}{4}(4-h^2).$

So, $y=\pm \frac{\sqrt{3}}{2}\sqrt{4-h^2}.$

Hence,

\qquad Q\left(h,-\frac{\sqrt{3}}{2}\sqrt{4-h^2}\right).$$ Let $$k=\frac{\sqrt{3}}{2}\sqrt{4-h^2}.$$ Then $$P(h,k),\quad Q(h,-k).$$ --- 2. **Equation of tangent at a point on the ellipse** For the ellipse $$\frac{x^2}{4}+\frac{y^2}{3}=1,$$ the tangent at $(x_1,y_1)$ is $$\frac{xx_1}{4}+\frac{yy_1}{3}=1.$$ So tangent at $P(h,k)$ is $$\frac{hx}{4}+\frac{ky}{3}=1. \tag{1}$$ Tangent at $Q(h,-k)$ is $$\frac{hx}{4}-\frac{ky}{3}=1. \tag{2}$$ --- 3. **Find point of intersection $R$ of the tangents** Add (1) and (2): $$\frac{hx}{2}=2 \quad \Rightarrow \quad hx=4 \quad \Rightarrow \quad x=\frac{4}{h}.$$ Subtract (2) from (1): $$\frac{2ky}{3}=0 \quad \Rightarrow \quad y=0.$$ Thus, $$R\left(\frac{4}{h},0\right).$$ --- 4. **Area of triangle $PQR$** Since $P$ and $Q$ lie on the vertical line $x=h$, the base $PQ$ has length $$PQ=2k=\sqrt{3}\sqrt{4-h^2}.$$ The perpendicular distance from $R\left(\frac{4}{h},0\right)$ to the line $x=h$ is $$\frac{4}{h}-h,$$ since $h\in [1/2,1]$ so $\frac{4}{h}>h$. Therefore, $$\Delta(h)=\frac12 \times PQ \times \left(\frac{4}{h}-h\right).$$ So, $$\Delta(h)=\frac12\cdot \sqrt{3}\sqrt{4-h^2}\left(\frac{4}{h}-h\right).$$ Now, $$\frac{4}{h}-h=\frac{4-h^2}{h}.$$ Hence, $$\Delta(h)=\frac{\sqrt{3}}{2}\sqrt{4-h^2}\cdot \frac{4-h^2}{h} =\frac{\sqrt{3}}{2}\frac{(4-h^2)^{3/2}}{h}.$$ --- 5. **Find maximum and minimum on $\left[\frac12,1\right]$** We need extrema of $$\Delta(h)=\frac{\sqrt{3}}{2}\frac{(4-h^2)^{3/2}}{h}.$$ The constant $\frac{\sqrt{3}}{2}$ does not affect monotonicity, so consider $$f(h)=\frac{(4-h^2)^{3/2}}{h}.$$ Differentiate: $$f(h)=(4-h^2)^{3/2}h^{-1}.$$ Using logarithmic differentiation, $$\ln f=\frac32\ln(4-h^2)-\ln h$$ $$\frac{f'}{f}=\frac32\cdot \frac{-2h}{4-h^2}-\frac1h = -\frac{3h}{4-h^2}-\frac1h.$$ Thus, $$\frac{f'}{f}= -\frac{3h^2+(4-h^2)}{h(4-h^2)} = -\frac{2h^2+4}{h(4-h^2)}<0$$ for all $h\in\left[\frac12,1\right]$. Hence $f(h)$, and therefore $\Delta(h)$, is strictly decreasing on $\left[\frac12,1\right]$. So, $$\Delta_1=\Delta\left(\frac12\right), \qquad \Delta_2=\Delta(1).$$ --- 6. **Compute $\Delta_1$ and $\Delta_2$** ### For $h=\frac12$: $$4-h^2=4-\frac14=\frac{15}{4}.$$ Then $$\Delta_1=\frac{\sqrt{3}}{2}\cdot \frac{\left(\frac{15}{4}\right)^{3/2}}{1/2} =\sqrt{3}\left(\frac{15}{4}\right)^{3/2}.$$ Now, $$\left(\frac{15}{4}\right)^{3/2}=\left(\frac{\sqrt{15}}{2}\right)^3=\frac{15\sqrt{15}}{8}.$$ So, $$\Delta_1=\sqrt{3}\cdot \frac{15\sqrt{15}}{8} =\frac{15\sqrt{45}}{8} =\frac{45\sqrt{5}}{8}.$$ ### For $h=1$: $$4-h^2=3.$$ Thus, $$\Delta_2=\frac{\sqrt{3}}{2}\cdot 3^{3/2} =\frac{\sqrt{3}}{2}\cdot 3\sqrt{3} =\frac{9}{2}.$$ --- 7. **Evaluate the required expression** We need $$\frac{8}{\sqrt{5}}\Delta_1-8\Delta_2.$$ Substitute: $$\frac{8}{\sqrt{5}}\cdot \frac{45\sqrt{5}}{8}-8\cdot \frac92 =45-36=9.$$ --- 8. **Final answer** $$\boxed{9}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: 9

Equation of the ellipse and points of intersection

The ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1.$

A vertical line through $(h,0)$ is $x=h.$

So, $y=\pm \frac{\sqrt{3}}{2}\sqrt{4-h^2}.$

Hence,

\qquad Q\left(h,-\frac{\sqrt{3}}{2}\sqrt{4-h^2}\right).$$ Let $$k=\frac{\sqrt{3}}{2}\sqrt{4-h^2}.$$ Then $$P(h,k),\quad Q(h,-k).$$ --- 2. **Equation of tangent at a point on the ellipse** For the ellipse $$\frac{x^2}{4}+\frac{y^2}{3}=1,$$ the tangent at $(x_1,y_1)$ is $$\frac{xx_1}{4}+\frac{yy_1}{3}=1.$$ So tangent at $P(h,k)$ is $$\frac{hx}{4}+\frac{ky}{3}=1. \tag{1}$$ Tangent at $Q(h,-k)$ is $$\frac{hx}{4}-\frac{ky}{3}=1. \tag{2}$$ --- 3. **Find point of intersection $R$ of the tangents** Add (1) and (2): $$\frac{hx}{2}=2 \quad \Rightarrow \quad hx=4 \quad \Rightarrow \quad x=\frac{4}{h}.$$ Subtract (2) from (1): $$\frac{2ky}{3}=0 \quad \Rightarrow \quad y=0.$$ Thus, $$R\left(\frac{4}{h},0\right).$$ --- 4. **Area of triangle $PQR$** Since $P$ and $Q$ lie on the vertical line $x=h$, the base $PQ$ has length $$PQ=2k=\sqrt{3}\sqrt{4-h^2}.$$ The perpendicular distance from $R\left(\frac{4}{h},0\right)$ to the line $x=h$ is $$\frac{4}{h}-h,$$ since $h\in [1/2,1]$ so $\frac{4}{h}>h$. Therefore, $$\Delta(h)=\frac12 \times PQ \times \left(\frac{4}{h}-h\right).$$ So, $$\Delta(h)=\frac12\cdot \sqrt{3}\sqrt{4-h^2}\left(\frac{4}{h}-h\right).$$ Now, $$\frac{4}{h}-h=\frac{4-h^2}{h}.$$ Hence, $$\Delta(h)=\frac{\sqrt{3}}{2}\sqrt{4-h^2}\cdot \frac{4-h^2}{h} =\frac{\sqrt{3}}{2}\frac{(4-h^2)^{3/2}}{h}.$$ --- 5. **Find maximum and minimum on $\left[\frac12,1\right]$** We need extrema of $$\Delta(h)=\frac{\sqrt{3}}{2}\frac{(4-h^2)^{3/2}}{h}.$$ The constant $\frac{\sqrt{3}}{2}$ does not affect monotonicity, so consider $$f(h)=\frac{(4-h^2)^{3/2}}{h}.$$ Differentiate: $$f(h)=(4-h^2)^{3/2}h^{-1}.$$ Using logarithmic differentiation, $$\ln f=\frac32\ln(4-h^2)-\ln h$$ $$\frac{f'}{f}=\frac32\cdot \frac{-2h}{4-h^2}-\frac1h = -\frac{3h}{4-h^2}-\frac1h.$$ Thus, $$\frac{f'}{f}= -\frac{3h^2+(4-h^2)}{h(4-h^2)} = -\frac{2h^2+4}{h(4-h^2)}<0$$ for all $h\in\left[\frac12,1\right]$. Hence $f(h)$, and therefore $\Delta(h)$, is strictly decreasing on $\left[\frac12,1\right]$. So, $$\Delta_1=\Delta\left(\frac12\right), \qquad \Delta_2=\Delta(1).$$ --- 6. **Compute $\Delta_1$ and $\Delta_2$** ### For $h=\frac12$: $$4-h^2=4-\frac14=\frac{15}{4}.$$ Then $$\Delta_1=\frac{\sqrt{3}}{2}\cdot \frac{\left(\frac{15}{4}\right)^{3/2}}{1/2} =\sqrt{3}\left(\frac{15}{4}\right)^{3/2}.$$ Now, $$\left(\frac{15}{4}\right)^{3/2}=\left(\frac{\sqrt{15}}{2}\right)^3=\frac{15\sqrt{15}}{8}.$$ So, $$\Delta_1=\sqrt{3}\cdot \frac{15\sqrt{15}}{8} =\frac{15\sqrt{45}}{8} =\frac{45\sqrt{5}}{8}.$$ ### For $h=1$: $$4-h^2=3.$$ Thus, $$\Delta_2=\frac{\sqrt{3}}{2}\cdot 3^{3/2} =\frac{\sqrt{3}}{2}\cdot 3\sqrt{3} =\frac{9}{2}.$$ --- 7. **Evaluate the required expression** We need $$\frac{8}{\sqrt{5}}\Delta_1-8\Delta_2.$$ Substitute: $$\frac{8}{\sqrt{5}}\cdot \frac{45\sqrt{5}}{8}-8\cdot \frac92 =45-36=9.$$ --- 8. **Final answer** $$\boxed{9}$$ This matches the stored correct answer.

Ellipse question

Ellipse question

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