Let \(x=2t,\ y=\frac {t^2}{3}\) be a conic. Let S be the focus and B be the point on the axis of the conic such that SA⊥BA, where A is any point on the conic. If k is the ordinate of the centroid of the ΔSAB, then \(\lim\limits_{t \to 1} k\) is equal to
\(\frac {17}{18}\)
\(\frac {19}{18}\)
\(\frac {11}{18}\)
\(\frac {13}{18}\)
The Correct Option is D
Solution and Explanation
\(x=2t,\ y=\frac 23\)
For \(t=1\)
\(A=(2,\frac 13)\)
Conic is \(x^2 = 12y \)\(⇒ S = (0, 3)\)
Let \(B = (0, β)\)
Given \(SA⊥BA\)
\((\frac {\frac 13}{2−3})(\frac {β−\frac 13}{−2})=−1\)
\((β−\frac 13)\frac 13=−2\)
\(β=\frac 13(−\frac {17}{3})\)
Ordinate of centroid,
\(K = \frac {β+\frac 13+3}{3}\)
\(=\frac {−\frac {17}{9}+\frac {10}{3}}{3}\)
\(= \frac {13}{18}\)
So, the correct option is (D): \(\frac {13}{18}\)
Top Questions on Conic sections
- Let the ellipse $ 3x^2 + py^2 = 4 $ pass through the centre $ C $ of the circle $ x^2 + y^2 - 2x - 4y - 11 = 0 $ of radius $ r $. Let $ f_1, f_2 $ be the focal distances of the point $ C $ on the ellipse. Then $ 6f_1 f_2 - r $ is equal to
- JEE Main - 2025
- Mathematics
- Conic sections
- Consider the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, $ having one of its foci at $ P(-3, 0) $. If the latus rectum through its other focus subtends a right angle at $ P $, and $ a^2b^2 = \alpha\sqrt{2} - \beta, \quad \alpha, \beta \in \mathbb{N}, $ then find $ \alpha $ and $ \beta $.
- JEE Main - 2025
- Mathematics
- Conic sections
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
- JEE Main - 2025
- Mathematics
- Conic sections
- The roots of the quadratic equation \( x^2 - 16 = 0 \) are:
- TS POLYCET - 2025
- Mathematics
- Conic sections
- If $ \theta $ is the angle subtended by a latus rectum at the center of the hyperbola having eccentricity $$ \frac{2}{\sqrt{7} - \sqrt{3}}, $$ then find $ \sin \theta $.
- AP EAPCET - 2025
- Mathematics
- Conic sections
Questions Asked in JEE Main exam
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- JEE Main - 2025
- Solutions
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.- JEE Main - 2025
- Electrical Circuits
- Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k} \). Let \( L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \vec{a}, \lambda \in {R} \) and \( L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu \vec{b}, \mu \in \mathbb{R} \) be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \vec{a} + \vec{b} \), then \( L_3 \) passes through the point:
- JEE Main - 2025
- Vectors
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
Concepts Used:
Conic Sections
When a plane intersects a cone in multiple sections, several types of curves are obtained. These curves can be a circle, an ellipse, a parabola, and a hyperbola. When a plane cuts the cone other than the vertex then the following situations may occur:
Let ‘β’ is the angle made by the plane with the vertical axis of the cone
- When β = 90°, we say the section is a circle
- When α < β < 90°, then the section is an ellipse
- When α = β; then the section is said to as a parabola
- When 0 ≤ β < α; then the section is said to as a hyperbola
Read More: Conic Sections