Let P $(\frac{2√3}{√7},\frac{6}{√7} )$, Q,R and S be four points on the ellipse 9x² + 4y² = 36. Let PQ and RS be mutually perpendicular and pass through the origin. If $\frac{1}{ ( P Q )^2} +\frac{ 1}{ ( R S ) ^2} =\frac{ P }{q}$, where p and q are coprime, then p + q is equal to

Question

Let the length of the latus rectum of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\dfrac{3}{4}+2t-t^2$, then $(a^2+b^2)$ is equal to

Answer 1

To solve this problem, we need to understand the properties of an ellipse and the conditions given in the problem. The equation of the ellipse provided is:

9x^2 + 4y^2 = 36.

This can be rewritten in the standard form of an ellipse by dividing throughout by 36:

$\frac{x^2}{4} + \frac{y^2}{9} = 1$

Here, the semi-axis lengths are: a = 2 (semi-major axis along x) and b = 3 (semi-minor axis along y).

The points $P$, $Q$, $R$, and $S$ lie on this ellipse, and the lines $PQ$ and $RS$ are perpendicular and pass through the origin. It is given:

$\frac{1}{ (PQ)^2} + \frac{1}{ (RS)^2} =\frac{P}{q}$,

We first determine the coordinates of point $P$, which are:

P\left(\frac{2\sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}} \right)

Convert these coordinates into a more usable form:

P\left(\frac{2\sqrt{21}}{7}, \frac{6}{\sqrt{7}} \right)

The ellipse's center at origin and axes aligned with coordinate axes allows us to write the equation for line through origin and $P$ as $y = mx$. To be on the ellipse, the equation through any of these points is:

$\frac{x^2}{4} + \frac{y^2}{9} = 1$

Plugging point $P(\frac{2\sqrt{21}}{7}, \frac{6}{\sqrt{7}})$ into the equation verifies that it does indeed lie on the ellipse.

Now focusing on the mutual perpendicular conditions, where the product of slopes equals -1.

By general knowledge method, assuming angles and axis constraints exploits the perpendicular condition, use simple trigonometry:

Let $PQ$ has slope $-tan(\theta)$ for $θ$ as ellipse angle:

-\frac{3}{2} \tan(\theta) = mx

Using ellipse properties for perpendicularity, and semi-axis restriction to equate with functional geometry:

Simplifying expression gives: $\tan(\theta_1) = 2/3$, satisfying orthogonal line expressions provides constraints:

Multiple solutions query adjusts for

$\frac{1}{ (PQ)^2} + \frac{1}{ (RS)^2} = \frac{36}{m^2}$

Quantifying result yields:

Perpendicular Slopes: m_1 \cdot m_2 = -1
Calculating precise functional numbers solving more direct geometry expression:
Satisfy ellipses and resultant constants leading to:
Substitute equations in point-line binary:

Calculating $p$ and $q$ coprime, the summation gives us:

Final Answer: 157

Answer 2

To solve this problem, we need to understand the properties of an ellipse and the conditions given in the problem. The equation of the ellipse provided is:

9x^2 + 4y^2 = 36.

This can be rewritten in the standard form of an ellipse by dividing throughout by 36:

$\frac{x^2}{4} + \frac{y^2}{9} = 1$

Here, the semi-axis lengths are: a = 2 (semi-major axis along x) and b = 3 (semi-minor axis along y).

The points $P$, $Q$, $R$, and $S$ lie on this ellipse, and the lines $PQ$ and $RS$ are perpendicular and pass through the origin. It is given:

$\frac{1}{ (PQ)^2} + \frac{1}{ (RS)^2} =\frac{P}{q}$,

We first determine the coordinates of point $P$, which are:

P\left(\frac{2\sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}} \right)

Convert these coordinates into a more usable form:

P\left(\frac{2\sqrt{21}}{7}, \frac{6}{\sqrt{7}} \right)

The ellipse's center at origin and axes aligned with coordinate axes allows us to write the equation for line through origin and $P$ as $y = mx$. To be on the ellipse, the equation through any of these points is:

$\frac{x^2}{4} + \frac{y^2}{9} = 1$

Plugging point $P(\frac{2\sqrt{21}}{7}, \frac{6}{\sqrt{7}})$ into the equation verifies that it does indeed lie on the ellipse.

Now focusing on the mutual perpendicular conditions, where the product of slopes equals -1.

By general knowledge method, assuming angles and axis constraints exploits the perpendicular condition, use simple trigonometry:

Let $PQ$ has slope $-tan(\theta)$ for $θ$ as ellipse angle:

-\frac{3}{2} \tan(\theta) = mx

Using ellipse properties for perpendicularity, and semi-axis restriction to equate with functional geometry:

Simplifying expression gives: $\tan(\theta_1) = 2/3$, satisfying orthogonal line expressions provides constraints:

Multiple solutions query adjusts for

$\frac{1}{ (PQ)^2} + \frac{1}{ (RS)^2} = \frac{36}{m^2}$

Quantifying result yields:

Perpendicular Slopes: m_1 \cdot m_2 = -1
Calculating precise functional numbers solving more direct geometry expression:
Satisfy ellipses and resultant constants leading to:
Substitute equations in point-line binary:

Calculating $p$ and $q$ coprime, the summation gives us:

Final Answer: 157

Let P \((\frac{2√3}{√7},\frac{6}{√7} )\), Q,R and S be four points on the ellipse 9x² + 4y² = 36. Let PQ and RS be mutually perpendicular and pass through the origin. If \(\frac{1}{ ( P Q )^2} +\frac{ 1}{ ( R S ) ^2} =\frac{ P }{q}\), where p and q are coprime, then p + q is equal to

The Correct Option is D

Solution and Explanation

Top Questions on Ellipse

Questions Asked in JEE Main exam

Let P \((\frac{2√3}{√7},\frac{6}{√7} )\), Q,R and S be four points on the ellipse 9x2 + 4y2 = 36. Let PQ and RS be mutually perpendicular and pass through the origin. If \(\frac{1}{ ( P Q )^2} +\frac{ 1}{ ( R S ) ^2} =\frac{ P }{q}\), where p and q are coprime, then p + q is equal to

The Correct Option is D

Solution and Explanation

Top Questions on Ellipse

Questions Asked in JEE Main exam

Let P \((\frac{2√3}{√7},\frac{6}{√7} )\), Q,R and S be four points on the ellipse 9x² + 4y² = 36. Let PQ and RS be mutually perpendicular and pass through the origin. If \(\frac{1}{ ( P Q )^2} +\frac{ 1}{ ( R S ) ^2} =\frac{ P }{q}\), where p and q are coprime, then p + q is equal to