Question:

The line $ y = 2x + c$ touches the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ if c is equal to

Updated On: Apr 18, 2024
  • 0
  • $\pm 2\sqrt{17}$
  • $c=\pm \sqrt{15}$
  • $c=\pm \sqrt{17}$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

$y= 2x+c$ touches $\frac{x^{2}}{16}+\frac{y^{2}}{4} =1$ if
$c^{2} = a^{2}m^{2}+b^{2} $
$ \quad\left({\text{Here}}\,\, a^{2}=16, b^{2}=4, m=2\right)$
i.e.,if $c^{2}=16\left(4\right)+4=68$
$\therefore c = \pm2\sqrt{17}$
Was this answer helpful?
0
0

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}