The line 12 x cos theta + 5 y sin theta = 60 | vedclass

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

Question

(B) The given line is $12 x \cos 	heta + 5 y \sin 	heta = 60$. Dividing by $60$,we get $\frac{12 x \cos 	heta}{60} + \frac{5 y \sin 	heta}{60} = 1

Vedclass · Accepted Answer

$144 x^{2} + 25 y^{2} = 3600$

Vedclass · Answer

(B) The given line is $12 x \cos 	heta + 5 y \sin 	heta = 60$. Dividing by $60$,we get $\frac{12 x \cos 	heta}{60} + \frac{5 y \sin 	heta}{60} = 1$,which simplifies to $\frac{x \cos 	heta}{5} + \frac{y \sin 	heta}{12} = 1$. Comparing this with the tangent equation of an ellipse $\frac{x}{a} \cos 	heta + \frac{y}{b} \sin 	heta = 1$,we identify $a = 5$ and $b = 12$. The equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,which is $\frac{x^{2}}{25} + \frac{y^{2}}{144} = 1$. Multiplying by $3600$,we get $144 x^{2} + 25 y^{2} = 3600$.

The line $12 x \cos \theta + 5 y \sin \theta = 60$ is tangent to which of the following curves?

Similar Questions

The eccentricity of the conic $4x^2 + 16y^2 - 24x - 3y = 1$ is

If the length of the minor axis of an ellipse is equal to one-fourth of the distance between the foci,then the eccentricity of the ellipse is:

$A$ focus of an ellipse having eccentricity $e = \frac{1}{2}$ is at $(0,0)$ and a directrix is the line $x = 4$. Then the equation of one such ellipse is

From point $P(8, 27)$,tangents $PQ$ and $PR$ are drawn to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$. The angle subtended by $QR$ at the origin is:

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module