Question:
If the normal at the point $ P(\theta) $ to the ellipse $ \frac{x^2}{14} + \frac {y^2}{5} = 1 $ intersects it again at the point $ Q(2\theta), $ then $ cos \theta $ equals to
If the normal at the point $ P(\theta) $ to the ellipse $ \frac{x^2}{14} + \frac {y^2}{5} = 1 $ intersects it again at the point $ Q(2\theta), $ then $ cos \theta $ equals to
Updated On: Jun 14, 2022
- $ -\frac{2}{3} $
- $ \frac{2}{3} $
- $ \frac{3}{2} $
- $ -\frac{3}{2} $
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Any point on the ellipse is
$(\sqrt{14} cos\,\theta, \sqrt{5} sin\,\theta)$
$\therefore$ Equation of normal at
$(\sqrt{14}\,cos\,\theta, \sqrt{5}\,sin\,\theta)$ is
$\sqrt{14} x\,sec\,\theta, \sqrt{5}y\,cosec\,\theta = 9$
Since, it passes through
$(\sqrt{14} \,cos\,2\theta, \sqrt{5}\,sin\,2\theta)$
$\therefore \sqrt{14}\sqrt{14} \,cos\,2\theta \,sec\,\theta$
$- \sqrt{5} \sqrt{5}\,sin\,2\theta \,cosec\,\theta = 9$
$\Rightarrow 14 \frac{cos\,2\theta}{cos\,\theta} - 5 \frac{sin\,\theta}{sin\,\theta} = 9$
$\Rightarrow 14(2\,cos^2\,\theta -1) - 10\,cos^2\,\theta = 9\,cos\,\theta$
$\Rightarrow 18\,cos^2\theta - 9\,cos\,\theta - 14 = 0$
$\Rightarrow (3\,cos\,\theta + 2)(6\,cos\,\theta - 7) = 0$
$\Rightarrow cos\,\theta = -\frac{2}{3}, cos\,\theta \ne \frac{7}{6}$
$(\sqrt{14} cos\,\theta, \sqrt{5} sin\,\theta)$
$\therefore$ Equation of normal at
$(\sqrt{14}\,cos\,\theta, \sqrt{5}\,sin\,\theta)$ is
$\sqrt{14} x\,sec\,\theta, \sqrt{5}y\,cosec\,\theta = 9$
Since, it passes through
$(\sqrt{14} \,cos\,2\theta, \sqrt{5}\,sin\,2\theta)$
$\therefore \sqrt{14}\sqrt{14} \,cos\,2\theta \,sec\,\theta$
$- \sqrt{5} \sqrt{5}\,sin\,2\theta \,cosec\,\theta = 9$
$\Rightarrow 14 \frac{cos\,2\theta}{cos\,\theta} - 5 \frac{sin\,\theta}{sin\,\theta} = 9$
$\Rightarrow 14(2\,cos^2\,\theta -1) - 10\,cos^2\,\theta = 9\,cos\,\theta$
$\Rightarrow 18\,cos^2\theta - 9\,cos\,\theta - 14 = 0$
$\Rightarrow (3\,cos\,\theta + 2)(6\,cos\,\theta - 7) = 0$
$\Rightarrow cos\,\theta = -\frac{2}{3}, cos\,\theta \ne \frac{7}{6}$
Was this answer helpful?
0
0
Top Questions on Ellipse
- Let the product of the focal distances of the point $$ \left( \sqrt{3}, \frac{1}{2} \right) $$ on the ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad (a > b), $$ be $ \frac{7}{4} $. Then the absolute difference of the eccentricities of two such ellipses is:
- If the equation of an ellipse \( E \) is \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \), then the length of the latus rectum of \( E \) is?
- Let the ellipse \( E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \( a > b \) and \( E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \), \( A < B \), have the same eccentricity \( \sqrt{\frac{1}{3}} \). Let the product of their lengths of latus rectums be \( \sqrt{\frac{32}{3}} \) and the distance between the foci of \( E_1 \) be 4. If \( E_1 \) and \( E_2 \) meet at \( A \), \( B \), \( C \), and \( D \), then the area of the quadrilateral ABCD equals:
- If \( 4x - 3y - 5 = 0 \) is a normal to the ellipse \( 3x^2 + 8y^2 = k \), then the equation of the tangent at point (-2,m) is:
- The area of the region enclosed by the curve $ \left\{ (x, y) : 4x^2 + 25y^2 = 100 \right\} $ is:
View More Questions
Questions Asked in AMUEEE exam
- The ratio of the half-life time $ (t_{1/2}) $ , to the three quarter life-time, $ (t_{3/4}) $ , for a reaction that is second order
- AMUEEE - 2018
- Order of Reaction
- $ 0.002\, M $ solution of a weak acid has an equivalent conductance $ (\wedge) 60\, ohm^{-1}\, cm^{2}\, eq^{-1} $ . What will be the $ pH $ ? (Given : $ \wedge = 400 \, ohm^{-1} \, cm^{2} \, eq^{-1} $ ]
- AMUEEE - 2018
- Electrochemistry
- The electrode potential, $ E^{\circ} $ , for the reduction of $ MnO^{-}_{4} $ to $ Mn^{2+} $ in acidic medium is $ +1.51\, V $ . Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for $ Zn^{2+}, Ag^{+} $ , and $ Au^{+} $ are given as $ Zn^{2+} (aq) +2e \to Zn(s), E^{\circ} = -0.762\, V $ $ Ag+ (aq) +e {\rightleftharpoons} Ag (s), E^{\circ}=+0.80 \,V $ $ Au^{+} (aq) +e {\rightleftharpoons} Au(s), E^{\circ}+1.69\,V $
- AMUEEE - 2018
- Electrochemistry
- Three particles, two with masses $ m $ and one with mass $ M $ , might be arranged in any of the four configurations shown below. Rank the configurations according to the magnitude of the gravitational force on $ M $ , least to greatest
- AMUEEE - 2018
- Gravitation
- The $ EAN $ value $ y \left[Ti\left(\sigma-C_{6}H_{5}\right)_{2}\left(\pi-C_{5}H_{5}\right)_{2}\right]^{0} $ is
- AMUEEE - 2018
- coordination compounds
View More Questions
Concepts Used:
Ellipse
Ellipse Shape
An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity
Properties
- Ellipse has two focal points, also called foci.
- The fixed distance is called a directrix.
- The eccentricity of the ellipse lies between 0 to 1. 0≤e<1
- The total sum of each distance from the locus of an ellipse to the two focal points is constant
- Ellipse has one major axis and one minor axis and a center
Read More: Conic Section
Eccentricity of the Ellipse
The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.
The eccentricity of ellipse, e = c/a
Where c is the focal length and a is length of the semi-major axis.
Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]
Area of an ellipse
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
Position of point related to Ellipse
Let the point p(x1, y1) and ellipse
(x2 / a2) + (y2 / b2) = 1
If [(x12 / a2)+ (y12 / b2) − 1)]
= 0 {on the curve}
<0{inside the curve}
>0 {outside the curve}