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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
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