If the equation of an ellipse is x216+y212=1 then

Mid point =(x1,y1) equation of chord is x1x16+y1y12−1=x1216+y1212−1Touches circle ∴25(x12256+y12144)=(x1216+y1212)2Normal condition a2l2+b2m2=(a2−b2)2n2 ⇒164+129=(16−12)2K2⇒K=±3Let foot of perpendicular =(x1,y1)⇒ slope of tangent =−x1y1 Equation is y−y1=−x1y1(x−x1)⇒x1x+y1y=x12+y12 ∴16x12+12y12=(x12+y12)2Equation of director circle is x2+y2=a2+b2=28

If the equation of an ellipse is x216+y212=1 then

Mid point =(x1,y1) equation of chord is x1x16+y1y12−1=x1216+y1212−1Touches circle ∴25(x12256+y12144)=(x1216+y1212)2Normal condition a2l2+b2m2=(a2−b2)2n2 ⇒164+129=(16−12)2K2⇒K=±3Let foot of perpendicular =(x1,y1)⇒ slope of tangent =−x1y1 Equation is y−y1=−x1y1(x−x1)⇒x1x+y1y=x12+y12 ∴16x12+12y12=(x12+y12)2Equation of director circle is x2+y2=a2+b2=28

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If the equation of an ellipse is $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The locus of middle points of chords of ellipse which touch the circle $x^{2} + y^{2} = 25$ is $25 (\frac{x^{2}}{256} + \frac{y^{2}}{144}) = {(\frac{x^{2}}{16} + \frac{y^{2}}{12})}^{2}$

If the line $2 x - 3 y + K = 0$ is a normal to the ellipse then $4 \sqrt{3} | K | = 12$ .

The locus of foot of perpendicular drawn from the centre of the ellipse on any of its tangents is ${(x^{2} + y^{2})}^{2} = 16 x^{2} + 12 y^{2}$

The equation of the director circle of the ellipse is $x^{2} + y^{2} = 16$ .

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Mid point $= (x_{1}, y_{1})$
equation of chord is $\frac{x_{1} x}{16} + \frac{y_{1} y}{12} - 1 = \frac{x_{1}^{2}}{16} + \frac{y_{1}^{2}}{12} - 1$
Touches circle $∴ 25 (\frac{x_{1}^{2}}{256} + \frac{y_{1}^{2}}{144}) = {(\frac{x_{1}^{2}}{16} + \frac{y_{1}^{2}}{12})}^{2}$
Normal condition $\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} = \frac{{(a^{2} - b^{2})}^{2}}{n^{2}}$
$\Rightarrow \frac{16}{4} + \frac{12}{9} = \frac{{(16 - 12)}^{2}}{K^{2}} \Rightarrow K = \pm \sqrt{3}$
Let foot of perpendicular $= (x_{1}, y_{1}) \Rightarrow$ slope of tangent $= \frac{- x_{1}}{y_{1}}$
Equation is $y - y_{1} = \frac{- x_{1}}{y_{1}} (x - x_{1}) \Rightarrow x_{1} x + y_{1} y = x_{1}^{2} + y_{1}^{2}$
$∴ 16 x_{1}^{2} + 12 y_{1}^{2} = {(x_{1}^{2} + y_{1}^{2})}^{2}$
Equation of director circle is $x^{2} + y^{2} = a^{2} + b^{2} = 28$