A line passing through the point P$(\sqrt{5}, \sqrt{5})$ intersects the ellipse $ \frac{x^2}{36} + \frac{y^2}{25} = 1 $ at A and B such that (PA).(PB) is maximum. Then 5(PA$^2$ + PB$^2$) is equal to :
Show Hint
- 218
- 377
- 290
- 338
The Correct Option is D
Approach Solution - 1
Approach Solution -2
The problem asks for the value of \( 5(\text{PA}^2 + \text{PB}^2) \) where A and B are the intersection points of a line passing through P\((\sqrt{5}, \sqrt{5})\) with the ellipse \( \frac{x^2}{36} + \frac{y^2}{25} = 1 \). The line is oriented such that the product of the distances, (PA)(PB), is maximized.
Concept Used:
To solve this problem, we use the parametric equation of a line passing through a given point. The intersection of this line with the ellipse results in a quadratic equation in terms of the distance parameter \(r\). The roots of this equation, \(r_1\) and \(r_2\), represent the distances PA and PB.
1. Parametric equation of a line: A line passing through \( P(x_1, y_1) \) can be represented as \( x = x_1 + r \cos \theta, y = y_1 + r \sin \theta \), where \(r\) is the distance from P to any point \((x, y)\) on the line, and \(\theta\) is the angle the line makes with the positive x-axis.
2. Intersection with Ellipse: Substituting these parametric coordinates into the ellipse equation yields a quadratic equation in \(r\), of the form \( Ar^2 + Br + C = 0 \).
3. Product of Roots: The product of the roots of this quadratic equation is \( r_1 r_2 = \frac{C}{A} \). Since the point P is inside the ellipse, the points A and B are on opposite sides of P along the line. Thus, one root will be positive and the other negative. The product of distances is \( \text{PA} \cdot \text{PB} = |r_1 r_2| = \left| \frac{C}{A} \right| \).
4. Maximizing the Product: To maximize the product of distances, we need to find the value of \(\theta\) that optimizes the expression \( \left| \frac{C}{A} \right| \).
Step-by-Step Solution:
Step 1: Set up the parametric equation of the line passing through P\((\sqrt{5}, \sqrt{5})\).
\[ x = \sqrt{5} + r \cos \theta, \quad y = \sqrt{5} + r \sin \theta \]Step 2: Substitute these coordinates into the ellipse equation \( \frac{x^2}{36} + \frac{y^2}{25} = 1 \).
\[ \frac{(\sqrt{5} + r \cos \theta)^2}{36} + \frac{(\sqrt{5} + r \sin \theta)^2}{25} = 1 \]Multiplying by the LCM, which is \(36 \times 25 = 900\), we get:
\[ 25(\sqrt{5} + r \cos \theta)^2 + 36(\sqrt{5} + r \sin \theta)^2 = 900 \]Step 3: Expand and rearrange the equation to form a quadratic in \(r\).
\[ 25(5 + 2\sqrt{5}r \cos \theta + r^2 \cos^2 \theta) + 36(5 + 2\sqrt{5}r \sin \theta + r^2 \sin^2 \theta) = 900 \] \[ (125 + 50\sqrt{5}r \cos \theta + 25r^2 \cos^2 \theta) + (180 + 72\sqrt{5}r \sin \theta + 36r^2 \sin^2 \theta) = 900 \]Group the terms by powers of \(r\):
\[ r^2(25 \cos^2 \theta + 36 \sin^2 \theta) + r(50\sqrt{5} \cos \theta + 72\sqrt{5} \sin \theta) + (125 + 180 - 900) = 0 \] \[ r^2(25 \cos^2 \theta + 36 \sin^2 \theta) + 2\sqrt{5}r(25 \cos \theta + 36 \sin \theta) - 595 = 0 \]Step 4: Find the expression for the product of distances (PA) \(\cdot\) (PB).
The product of the roots \( r_1 r_2 \) is \( \frac{C}{A} \). The product of the distances is \( |\frac{C}{A}| \).
\[ \text{PA} \cdot \text{PB} = \left| \frac{-595}{25 \cos^2 \theta + 36 \sin^2 \theta} \right| = \frac{595}{25 \cos^2 \theta + 36 \sin^2 \theta} \]Step 5: Maximize the product (PA) \(\cdot\) (PB).
To maximize this expression, we need to minimize the denominator. Let's rewrite the denominator:
\[ D(\theta) = 25 \cos^2 \theta + 36 \sin^2 \theta = 25(1 - \sin^2 \theta) + 36 \sin^2 \theta = 25 + 11 \sin^2 \theta \]The minimum value of \( D(\theta) \) occurs when \( \sin^2 \theta \) is minimum, which is \( \sin^2 \theta = 0 \). This happens when \( \theta = 0 \) or \( \pi \), meaning the line is horizontal.
Step 6: Find the intersection points A and B for the horizontal line \( y = \sqrt{5} \).
Substitute \( y = \sqrt{5} \) into the ellipse equation:
\[ \frac{x^2}{36} + \frac{(\sqrt{5})^2}{25} = 1 \implies \frac{x^2}{36} + \frac{5}{25} = 1 \implies \frac{x^2}{36} = 1 - \frac{1}{5} = \frac{4}{5} \] \[ x^2 = \frac{144}{5} \implies x = \pm \frac{12}{\sqrt{5}} = \pm \frac{12\sqrt{5}}{5} \]The intersection points are \( A\left(\frac{12\sqrt{5}}{5}, \sqrt{5}\right) \) and \( B\left(-\frac{12\sqrt{5}}{5}, \sqrt{5}\right) \).
Step 7: Calculate the distances PA and PB.
The point is P\((\sqrt{5}, \sqrt{5})\).
\[ \text{PA} = \left| \frac{12\sqrt{5}}{5} - \sqrt{5} \right| = \left| \frac{12\sqrt{5} - 5\sqrt{5}}{5} \right| = \frac{7\sqrt{5}}{5} \] \[ \text{PB} = \left| -\frac{12\sqrt{5}}{5} - \sqrt{5} \right| = \left| \frac{-12\sqrt{5} - 5\sqrt{5}}{5} \right| = \frac{17\sqrt{5}}{5} \]Final Computation & Result:
Now, we compute the required expression \( 5(\text{PA}^2 + \text{PB}^2) \).
\[ \text{PA}^2 = \left(\frac{7\sqrt{5}}{5}\right)^2 = \frac{49 \times 5}{25} = \frac{49}{5} \] \[ \text{PB}^2 = \left(\frac{17\sqrt{5}}{5}\right)^2 = \frac{289 \times 5}{25} = \frac{289}{5} \] \[ \text{PA}^2 + \text{PB}^2 = \frac{49}{5} + \frac{289}{5} = \frac{338}{5} \]Finally, we multiply by 5:
\[ 5(\text{PA}^2 + \text{PB}^2) = 5 \times \frac{338}{5} = 338 \]The value of \( 5(\text{PA}^2 + \text{PB}^2) \) is 338.
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Coordinate Geometry Questions
- The area (in square units) of the region bounded by the parabola \(y^2 = 4(x - 2)\) and the line \(y = 2x - 8\) is:
In a △ABC, suppose y = x is the equation of the bisector of the angle B and the equation of the side AC is 2x−y = 2. If 2AB = BC and the points A and B are respectively (4, 6) and (α, β), then α + 2β is equal to:
- Let\((5,\frac{a}{4})\),be the circumcenter of a triangle with vertices A (a, -2),B (a, 6)and C \((\frac{a}{4},-2)\).Let \(\alpha\) denote the circumradius, \(\beta\) denote the area and \(\gamma\) denote the perimeter of the triangle. Then \(\alpha+\beta+\gamma\) is
- The area (in square units) of the part of the circle \(x^2 + y^2 = 169\) which is below the line \( 5x - y = 13 \) is\[\frac{\pi \alpha}{2 \beta} - \frac{65}{2} + \frac{\alpha}{\beta} \sin^{-1} \left( \frac{12}{13} \right)\]where \( \alpha \) and \( \beta \) are coprime numbers. Then \( \alpha + \beta \) is equal to
- The length of the chord of the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\), whose mid-point is \(\left(1, \frac{2}{5}\right)\), is equal to:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,