Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
the normal to the curve at P(x,y) meets the x axis at G. If the distance of G from the origin is twice the abscissa of P , then the curve is a) ellipse b) parabola c) ellipse or hyperbola the normal to the curve at P(x,y) meets the x axis at G. If the distance of G from the origin is twice the abscissa of P , then the curve is a) ellipseb) parabola c) ellipse or hyperbola
solution:for any curve the equation of normal at ( X1, Y1)(y- y1) = -1/m (x- X1) {where m = slope of tengent or say dy/dx at (x1, Y1) }normal meets the X axis at G..so ( 0 - y1) = -1/m (x- X1)so x = m*Y1 + X1so coordinates of G( m*Y1 + X1 , 0)now ...distance of G from the origin is twice the abscissa of Pso2X1 = m*Y1 + X1X1 = m*Y1m = X1/Y1dy/dx (at X1, Y1) = X1/Y1dy/dx = x/yydy = x dxintegrate both sidesy2/2 = x2/2 + cx2/2 - y2/2 = cso hyperbola..Thanks and Regards,Ajay verma,askIITians faculty,IIT HYDERABAD
thank you
but sir the answer says it can be both ellipse as well as hyperbola, so can u please prove this as an ellipse too.
The equation is |x1 + my1|=2|x1|So there are 2 equations possible one with positive and the other with negative
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -