120538 Let the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{7}=1\). Then the length of the latus rectum of the hyperbola is :

120539 If the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) meets the line \(\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis and the line \(\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis, then the eccentricity of the ellipse is
#[Qdiff: Hard, QCat: Numerical Based, examname: APEAPCET-20.08.2021,Shift-I], 787. The equation of an ellipse in its standard form, given the distance between its foci is 2 units and the length of its latus rectum is \(\frac{15}{2}\) units is,

120540 The equations of the latusrectum of the ellipse \(9 x^2+4 y^2-18 x-8 y-23=0\) are

120541 If \(\tan \theta_1 \cdot \tan \theta_2=\frac{-a^2}{b^2}\) then the chord joining 2 points \(\theta_1\) and \(\theta_2\) one the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) will subtend a right angle at

120542 If the lines \(2 x-y+3=0\) and \(4 x+k y+3=0\) are conjugate with respect to the ellipse \(5 x^2+\) \(6 y^2-15=0\), then ' \(k\) ' equals

120538 Let the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{7}=1\). Then the length of the latus rectum of the hyperbola is :

120539 If the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) meets the line \(\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis and the line \(\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis, then the eccentricity of the ellipse is
#[Qdiff: Hard, QCat: Numerical Based, examname: APEAPCET-20.08.2021,Shift-I], 787. The equation of an ellipse in its standard form, given the distance between its foci is 2 units and the length of its latus rectum is \(\frac{15}{2}\) units is,

120540 The equations of the latusrectum of the ellipse \(9 x^2+4 y^2-18 x-8 y-23=0\) are

120541 If \(\tan \theta_1 \cdot \tan \theta_2=\frac{-a^2}{b^2}\) then the chord joining 2 points \(\theta_1\) and \(\theta_2\) one the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) will subtend a right angle at

120542 If the lines \(2 x-y+3=0\) and \(4 x+k y+3=0\) are conjugate with respect to the ellipse \(5 x^2+\) \(6 y^2-15=0\), then ' \(k\) ' equals

120538 Let the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{7}=1\). Then the length of the latus rectum of the hyperbola is :

120539 If the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) meets the line \(\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis and the line \(\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis, then the eccentricity of the ellipse is
#[Qdiff: Hard, QCat: Numerical Based, examname: APEAPCET-20.08.2021,Shift-I], 787. The equation of an ellipse in its standard form, given the distance between its foci is 2 units and the length of its latus rectum is \(\frac{15}{2}\) units is,

120540 The equations of the latusrectum of the ellipse \(9 x^2+4 y^2-18 x-8 y-23=0\) are

120541 If \(\tan \theta_1 \cdot \tan \theta_2=\frac{-a^2}{b^2}\) then the chord joining 2 points \(\theta_1\) and \(\theta_2\) one the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) will subtend a right angle at

120542 If the lines \(2 x-y+3=0\) and \(4 x+k y+3=0\) are conjugate with respect to the ellipse \(5 x^2+\) \(6 y^2-15=0\), then ' \(k\) ' equals

120538 Let the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{7}=1\). Then the length of the latus rectum of the hyperbola is :

120539 If the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) meets the line \(\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis and the line \(\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis, then the eccentricity of the ellipse is
#[Qdiff: Hard, QCat: Numerical Based, examname: APEAPCET-20.08.2021,Shift-I], 787. The equation of an ellipse in its standard form, given the distance between its foci is 2 units and the length of its latus rectum is \(\frac{15}{2}\) units is,

120540 The equations of the latusrectum of the ellipse \(9 x^2+4 y^2-18 x-8 y-23=0\) are

120541 If \(\tan \theta_1 \cdot \tan \theta_2=\frac{-a^2}{b^2}\) then the chord joining 2 points \(\theta_1\) and \(\theta_2\) one the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) will subtend a right angle at

120542 If the lines \(2 x-y+3=0\) and \(4 x+k y+3=0\) are conjugate with respect to the ellipse \(5 x^2+\) \(6 y^2-15=0\), then ' \(k\) ' equals

120538 Let the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{7}=1\). Then the length of the latus rectum of the hyperbola is :

120539 If the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) meets the line \(\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis and the line \(\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1\) on the \(x\)-axis, then the eccentricity of the ellipse is
#[Qdiff: Hard, QCat: Numerical Based, examname: APEAPCET-20.08.2021,Shift-I], 787. The equation of an ellipse in its standard form, given the distance between its foci is 2 units and the length of its latus rectum is \(\frac{15}{2}\) units is,

120540 The equations of the latusrectum of the ellipse \(9 x^2+4 y^2-18 x-8 y-23=0\) are

120541 If \(\tan \theta_1 \cdot \tan \theta_2=\frac{-a^2}{b^2}\) then the chord joining 2 points \(\theta_1\) and \(\theta_2\) one the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) will subtend a right angle at

120542 If the lines \(2 x-y+3=0\) and \(4 x+k y+3=0\) are conjugate with respect to the ellipse \(5 x^2+\) \(6 y^2-15=0\), then ' \(k\) ' equals