Find the equations of the lines with equal intercepts on the axes & which touch the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

x + y - 5 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x - y - 5 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + y + 5 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x - y + 5 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $x + y - 5 = 0$
C $x + y + 5 = 0$
Let the equation of line be $x + y = k$
Here, $m = - 1$ and $c = k$
Condition of tangency: $k = \pm \sqrt{a^{2} m^{2} + b^{2}} = \pm \sqrt{16 + 9} = \pm 5$

Suggest Corrections

Similar questions

Q. The equation of the circle which touches the lines

3 x + 4 y - 5 = 0

and

3 x + 4 y + 25 = 0

and whose centre lies on the line

x + 2 y = 0

, is

Q. The equation of the line which is perpendicular to

x + 4 y - 5 = 0

at it's

y

intercept

Find the equation of the lines through the point of intersection of the lines x-3y+1 = 0 and 2x+5y-9 = 0 and whose distance from the origin is $\sqrt{5}$ .

Q. The equation of line passing through

(3, 4)

and parallel to

5 x + 9 y + 12 = 0

Q. The equation of line perpendicular to

3 x + 5 y = 19

and passing through

(3, 2)

Join BYJU'S Learning Program

Find the equations of the lines with equal intercepts on the axes & which touch the ellipse x216+y29=1

The correct options are A x+y−5=0 C x+y+5=0Let the equation of line be x+y=kHere, m=−1 and c=kCondition of tangency: k=±√a2m2+b2=±√16+9=±5

Find the equations of the lines with equal intercepts on the axes & which touch the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

The correct options are
A $x + y - 5 = 0$
C $x + y + 5 = 0$
Let the equation of line be $x + y = k$
Here, $m = - 1$ and $c = k$
Condition of tangency: $k = \pm \sqrt{a^{2} m^{2} + b^{2}} = \pm \sqrt{16 + 9} = \pm 5$