Question:
Let \(\mathrm{P}\) be a point in the first quadrant lying on the ellipse \(9x^{2}+16y^{2}=144\), such that the tangent at \(\mathrm{P}\) to the ellipse is inclined at an angle \(135^{\circ}\) to the positive direction of \(x\)-axis. Then the coordinates of \(\mathrm{P}\) are :
Let \(\mathrm{P}\) be a point in the first quadrant lying on the ellipse \(9x^{2}+16y^{2}=144\), such that the tangent at \(\mathrm{P}\) to the ellipse is inclined at an angle \(135^{\circ}\) to the positive direction of \(x\)-axis. Then the coordinates of \(\mathrm{P}\) are :
Show Hint
Differentiate the ellipse equation to get the tangent slope, set it equal to \(\tan 135^{\circ}=-1\), and combine with the ellipse equation.
Updated On: Nov 12, 2026
- \(\left(\frac{\sqrt{143}}{3},\frac{1}{4}\right)\)
- \(\left(\frac{8}{9},\frac{\sqrt{77}}{3}\right)\)
- \(\left(\frac{4}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)\)
- \(\left(\frac{16}{5},\frac{9}{5}\right)\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Differentiating \(9x^{2}+16y^{2}=144\): \(18x+32y\,\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{9x}{16y}\).
Tangent slope \(=\tan135^{\circ}=-1\Rightarrow 9x=16y\Rightarrow x=\frac{16y}{9}\).
Substituting: \(9\cdot\frac{256y^{2}}{81}+16y^{2}=144\Rightarrow \frac{256y^{2}}{9}+\frac{144y^{2}}{9}=144\Rightarrow y^{2}=\frac{81}{25}\).
First quadrant: \(y=\frac{9}{5}\), \(x=\frac{16}{5}\).
Answer: (4) \(\left(\frac{16}{5},\frac{9}{5}\right)\)
Tangent slope \(=\tan135^{\circ}=-1\Rightarrow 9x=16y\Rightarrow x=\frac{16y}{9}\).
Substituting: \(9\cdot\frac{256y^{2}}{81}+16y^{2}=144\Rightarrow \frac{256y^{2}}{9}+\frac{144y^{2}}{9}=144\Rightarrow y^{2}=\frac{81}{25}\).
First quadrant: \(y=\frac{9}{5}\), \(x=\frac{16}{5}\).
Answer: (4) \(\left(\frac{16}{5},\frac{9}{5}\right)\)
Was this answer helpful?
0
0
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_____
View More Questions
Top JEE Main Ellipse Questions
- If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:
- If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is.
- A parabola has vertex \((2, 3)\), equation of directrix is \(2x – y = 1\) and equation of ellipse is \(\frac {x^2}{a^2} +\frac {y^2}{b^2} =1\), \(e=\frac {1}{\sqrt 2}\) and ellipse passing through focur of parabola then square of length of latus rectum of ellipse is
- Let \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a>b \), be an ellipse whose eccentricity is \( \frac{1}{\sqrt{2}} \) and the length of the latus rectum is \( \sqrt{14} \). Then the square of the eccentricity of \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is:
- Let \( f(x) = x^2 + 9 \), \( g(x) = \frac{x}{x-9} \), and \[ a = f \circ g(10), \, b = g \circ f(3). \]
If \( e \) and \( l \) denote the eccentricity and the length of the latus rectum of the ellipse \[ \frac{x^2}{a} + \frac{y^2}{b} = 1, \] then \( 8e^2 + l^2 \) is equal to:
View More Questions
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,
View More Questions