Question:
If the distance of a point P on an ellipse from its focus (1, 2) is half of the distance of P from its corresponding directrix \( x+y=0 \), then the point of intersection of the given directrix and its major axis, is
If the distance of a point P on an ellipse from its focus (1, 2) is half of the distance of P from its corresponding directrix \( x+y=0 \), then the point of intersection of the given directrix and its major axis, is
Show Hint
The eccentricity value (\( e = 1/2 \)) given in the text is extra data. The axis direction depends entirely on the orientation of the directrix line and the position of the focus point.
Updated On: Jun 8, 2026
- \( (2,-2) \)
- \( (-\frac{1}{2},\frac{1}{2}) \)
- \( (-1,1) \)
- \( (\frac{1}{3},-\frac{1}{3}) \)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
By definition, the major axis of an ellipse is a line that passes through its focus and is strictly perpendicular to its directrix line. The intersection point can be found by writing the equation of the major axis line and solving it simultaneously with the directrix equation.
Step 1: Finding the slope of the major axis.
The given directrix line is: \[ x + y = 0 \implies y = -x \] The slope of the directrix is \( m_1 = -1 \). Since the major axis is perpendicular to the directrix, its slope \( m_2 \) must satisfy: \[ m_2 \times (-1) = -1 \implies m_2 = 1 \]
Step 2: Formulating the equation of the major axis.
The major axis passes through the focus point \( (1, 2) \) with a slope of 1: \[ y - 2 = 1(x - 1) \implies y - 2 = x - 1 \implies x - y + 1 = 0 \]
Step 3: Finding the intersection point.
We solve the system of linear equations: 1) \( x + y = 0 \implies y = -x \)
2) \( x - y + 1 = 0 \)
Substitute (1) into (2): \[ x - (-x) + 1 = 0 \implies 2x + 1 = 0 \implies x = -\frac{1}{2} \] This gives \( y = -(-\frac{1}{2}) = \frac{1}{2} \). Thus, the intersection point is \( \left(-\frac{1}{2}, \frac{1}{2}\right) \). This tracks Option (B).
Step 1: Finding the slope of the major axis.
The given directrix line is: \[ x + y = 0 \implies y = -x \] The slope of the directrix is \( m_1 = -1 \). Since the major axis is perpendicular to the directrix, its slope \( m_2 \) must satisfy: \[ m_2 \times (-1) = -1 \implies m_2 = 1 \]
Step 2: Formulating the equation of the major axis.
The major axis passes through the focus point \( (1, 2) \) with a slope of 1: \[ y - 2 = 1(x - 1) \implies y - 2 = x - 1 \implies x - y + 1 = 0 \]
Step 3: Finding the intersection point.
We solve the system of linear equations: 1) \( x + y = 0 \implies y = -x \)
2) \( x - y + 1 = 0 \)
Substitute (1) into (2): \[ x - (-x) + 1 = 0 \implies 2x + 1 = 0 \implies x = -\frac{1}{2} \] This gives \( y = -(-\frac{1}{2}) = \frac{1}{2} \). Thus, the intersection point is \( \left(-\frac{1}{2}, \frac{1}{2}\right) \). This tracks Option (B).
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET sections of a cone Questions
- The equation of the parabola with focus at $(3, 0)$ and directrix $x + 3 = 0$ is:
- If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is:
- The line $y = mx + c$ touches the ellipse $9x^2 + 16y^2 = 144$ if the value of $c^2$ is:
- If the eccentricity of a hyperbola is $\sqrt{3}$, then the eccentricity of its conjugate hyperbola is:
- The equation of the tangent to the parabola $y^2 = 8x$ which is parallel to the line $2x - y + 5 = 0$ is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions