If the lines 2x - y + 3 = 0 and 4x + ky + 3 = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given lines are $2x - y + 3 = 0$ and $4x + ky + 3 = 0$. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,two lines $a_1x + b_1y + c_1 =

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The given lines are $2x - y + 3 = 0$ and $4x + ky + 3 = 0$. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are conjugate if $a^2a_1a_2 + b^2b_1b_2 = c_1c_2$. Rewriting the ellipse equation $5x^2 + 6y^2 = 15$ as $\frac{x^2}{3} + \frac{y^2}{2.5} = 1$,we have $a^2 = 3$ and $b^2 = 2.5 = \frac{5}{2}$. Here,$a_1 = 2, b_1 = -1, c_1 = 3$ and $a_2 = 4, b_2 = k, c_2 = 3$. Substituting these values into the condition: $3(2)(4) + (2.5)(-1)(k) = (3)(3)$. $24 - 2.5k = 9$. $2.5k = 15$. $k = \frac{15}{2.5} = 6$.

If the lines $2x - y + 3 = 0$ and $4x + ky + 3 = 0$ are conjugate with respect to the ellipse $5x^2 + 6y^2 - 15 = 0$,then $k$ equals:

Similar Questions

The equation of the ellipse whose latus rectum is $8$ and whose eccentricity is $\frac{1}{\sqrt{2}}$,referred to the principal axes of coordinates,is

The eccentricity of the conic $16x^2 + 7y^2 = 112$ is

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$,foci $(\pm 4, 0)$.

Let a focus of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $S(4, 0)$ and its eccentricity be $\frac{4}{5}$. If the point $P(3, \alpha)$ lies on $E$ and $O$ is the origin, then the area of $\triangle POS$ is equal to: (in $/ 5$)

The locus of the midpoints of the intercepted portion of the tangents by the coordinate axes,which are drawn to the ellipse $x^2+2y^2=2$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module