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Show that the curves (x^2)/(a^2+lambda1)...

Show that the curves `(x^2)/(a^2+lambda_1)+(y^2)/(b^2+lambda_1)=1` and `(x^2)/(a^2+lambda_2)+(y^2)/(b^2+lambda_2)=1` intersect at right angles.

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To show that the curves \[ \frac{x^2}{a^2 + \lambda_1} + \frac{y^2}{b^2 + \lambda_1} = 1 \] and ...
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Tangents,one to each of the ellipses (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1 are drawn Tangents,If the tangents meet at right angles then the locus of their point of intersection is

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Knowledge Check

  • If (x^(2))/(lambda+3)+(y^(2))/(2-lambda)=1 represents a hyperbola then

    A
    `lambda in (2,3)`
    B
    `lambda in (-2,3)`
    C
    `lambda in (-3,2)`
    D
    `lambda in (2,infty)`

  • The locus of the point of intersection of perpendicular tangents to x^(2)/a^(2) + y^(2)/b^(2) = 1 and (x^(2))/(a^(2) + lambda) + (y^(2))/(b^(2) + lambda) = 1 , is

    A
    `x^(2) + y^(2) = a^(2) + lambda`
    B
    `x^(2) + y^(2) = b^(2) + lambda`
    C
    `x^(2) + y^(2) = a^(2) + b^(2) + lambda`
    D
    `x^(2) + y^(2) = a^(2) + b^(2)`

  • If the lines (x-1)/(1)=(y-3)/(1)=(z-2)/(lambda) and (x-1)/(lambda)=(y-3)/(2)=(z-4)/(1) intersect at a point, then the value of lambda^(2)+4 is equal to

    A
    8
    B
    10
    C
    13
    D
    5

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    Prove that the family of curves x^2/(a^2+lambda)+y^2/(b^2+lambda)=1 , where lambda is a parameter, is self orthogonal.

    Let alpha and beta be the values of x obtained form the equation lambda^(2) (x^(2)-x) 2lambdax +3 =0 and if lambda_(1),lambda^(2) be the two values of lambda for which alpha and beta are connected by the relation alpha/beta + beta/alpha = 4/3 . then find the value of (lambda_(1)^(2))/(lambda_(2)) + (lambda_(1)^(2))/(lambda_(1)) and (lambda_(1)^(2))/lambda_(2)^(2) + (lambda_(2)^(2))/(lambda_(1)^(2))

    If y=mx+c be a tangent to the hyperbola (x^(2))/(lambda^(2))-(y^(2))/((lambda^(3)+lambda^(2)+lambda)^(2))=1,(lambda!=0), then

    The differential equation satisfying the curve (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1 when lambda begin arbitary uknowm, is (a) (x+yy_(1))(xy_(1)-y)=(a^(2)-b^(2))y_(1) (b) (x+yy_(1))(xy_(1)-y)=y_(1) ( c ) (x-yy_(1))(xy_(1)+y)(a^(2)-b^(2))y_(1) (d) None of these

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