120642 If the tangent at a point \(P\) one the parabola \(y^2=\) \(3 x\) is parallel to the line \(x+2 y=1\) and tangents at the point \(Q\) and \(R\) on the ellipse \(\frac{x^2}{4}+\frac{y^2}{1}=1\) are perpendicular to the line \(x-y=\) 2, then the area of the triangle \(P Q R\) is:

120643 Let a circle of radius 4 be concentric to the ellipse \(15 x^2+19 y^2=285\). Then the common tangents are inclined to the minor axis of the ellipse at the angle.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-21.04.2019,Shift-I], So, Hence, Where, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), On comparing standard ellipse equation, \(\mathrm{a}^2=19, \mathrm{~b}^2=15\), Equation of tangent of ellipse is -, \(\mathrm{y}=\mathrm{mx}+\sqrt{19 \mathrm{~m}^2+15}\), \(\mathrm{mx}-\mathrm{y} \pm \sqrt{19 \mathrm{~m}^2+15}=0\), Perpendicular distance from centre of circle to tangent \(=\), 4, \(\left|\frac{ \pm \sqrt{19 \mathrm{~m}^2+15}}{\sqrt{\mathrm{m}^2+1}}\right|=4\), \(19 \mathrm{~m}^2+15=16 \mathrm{~m}^2+16\), \(\mathrm{~m}= \pm \frac{1}{\sqrt{3}}\), \(\tan \theta=\mathrm{m}=\frac{1}{\sqrt{3}}\), \(\theta=\frac{\pi}{6}\) with \(\mathrm{x}\)-axis or \(\frac{\pi}{3}\) with \(\mathrm{y}\)-axis, the common tangents are inclined to the minor axis of the ellipse at an angle of \(\frac{\pi}{3}\)., 911. If the tangent drawn to the parabola \(y^2=4 x\) at \(\left(t^2, 2 t\right)\) is the normal to the ellipse \(4 x^2+5 y^2=20\) at \((\sqrt{5} \cos \theta, 2 \sin \theta)\), then,

120644 The value of ' \(k\) ' so that the line \(y=2 x+k\) may touch the ellipse \(3 x^2+5 y^2=15\) is

120645 If the tangent at the point \((1,2)\) on the ellipse \(3 x^2+4 y^2=19\) is also a tangent to the parabola \(\mathbf{y}^2-\mathbf{k x}=\mathbf{0}\), then \(\mathbf{k}=\)

120642 If the tangent at a point \(P\) one the parabola \(y^2=\) \(3 x\) is parallel to the line \(x+2 y=1\) and tangents at the point \(Q\) and \(R\) on the ellipse \(\frac{x^2}{4}+\frac{y^2}{1}=1\) are perpendicular to the line \(x-y=\) 2, then the area of the triangle \(P Q R\) is:

120643 Let a circle of radius 4 be concentric to the ellipse \(15 x^2+19 y^2=285\). Then the common tangents are inclined to the minor axis of the ellipse at the angle.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-21.04.2019,Shift-I], So, Hence, Where, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), On comparing standard ellipse equation, \(\mathrm{a}^2=19, \mathrm{~b}^2=15\), Equation of tangent of ellipse is -, \(\mathrm{y}=\mathrm{mx}+\sqrt{19 \mathrm{~m}^2+15}\), \(\mathrm{mx}-\mathrm{y} \pm \sqrt{19 \mathrm{~m}^2+15}=0\), Perpendicular distance from centre of circle to tangent \(=\), 4, \(\left|\frac{ \pm \sqrt{19 \mathrm{~m}^2+15}}{\sqrt{\mathrm{m}^2+1}}\right|=4\), \(19 \mathrm{~m}^2+15=16 \mathrm{~m}^2+16\), \(\mathrm{~m}= \pm \frac{1}{\sqrt{3}}\), \(\tan \theta=\mathrm{m}=\frac{1}{\sqrt{3}}\), \(\theta=\frac{\pi}{6}\) with \(\mathrm{x}\)-axis or \(\frac{\pi}{3}\) with \(\mathrm{y}\)-axis, the common tangents are inclined to the minor axis of the ellipse at an angle of \(\frac{\pi}{3}\)., 911. If the tangent drawn to the parabola \(y^2=4 x\) at \(\left(t^2, 2 t\right)\) is the normal to the ellipse \(4 x^2+5 y^2=20\) at \((\sqrt{5} \cos \theta, 2 \sin \theta)\), then,

120644 The value of ' \(k\) ' so that the line \(y=2 x+k\) may touch the ellipse \(3 x^2+5 y^2=15\) is

120645 If the tangent at the point \((1,2)\) on the ellipse \(3 x^2+4 y^2=19\) is also a tangent to the parabola \(\mathbf{y}^2-\mathbf{k x}=\mathbf{0}\), then \(\mathbf{k}=\)

120642 If the tangent at a point \(P\) one the parabola \(y^2=\) \(3 x\) is parallel to the line \(x+2 y=1\) and tangents at the point \(Q\) and \(R\) on the ellipse \(\frac{x^2}{4}+\frac{y^2}{1}=1\) are perpendicular to the line \(x-y=\) 2, then the area of the triangle \(P Q R\) is:

120643 Let a circle of radius 4 be concentric to the ellipse \(15 x^2+19 y^2=285\). Then the common tangents are inclined to the minor axis of the ellipse at the angle.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-21.04.2019,Shift-I], So, Hence, Where, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), On comparing standard ellipse equation, \(\mathrm{a}^2=19, \mathrm{~b}^2=15\), Equation of tangent of ellipse is -, \(\mathrm{y}=\mathrm{mx}+\sqrt{19 \mathrm{~m}^2+15}\), \(\mathrm{mx}-\mathrm{y} \pm \sqrt{19 \mathrm{~m}^2+15}=0\), Perpendicular distance from centre of circle to tangent \(=\), 4, \(\left|\frac{ \pm \sqrt{19 \mathrm{~m}^2+15}}{\sqrt{\mathrm{m}^2+1}}\right|=4\), \(19 \mathrm{~m}^2+15=16 \mathrm{~m}^2+16\), \(\mathrm{~m}= \pm \frac{1}{\sqrt{3}}\), \(\tan \theta=\mathrm{m}=\frac{1}{\sqrt{3}}\), \(\theta=\frac{\pi}{6}\) with \(\mathrm{x}\)-axis or \(\frac{\pi}{3}\) with \(\mathrm{y}\)-axis, the common tangents are inclined to the minor axis of the ellipse at an angle of \(\frac{\pi}{3}\)., 911. If the tangent drawn to the parabola \(y^2=4 x\) at \(\left(t^2, 2 t\right)\) is the normal to the ellipse \(4 x^2+5 y^2=20\) at \((\sqrt{5} \cos \theta, 2 \sin \theta)\), then,

120644 The value of ' \(k\) ' so that the line \(y=2 x+k\) may touch the ellipse \(3 x^2+5 y^2=15\) is

120645 If the tangent at the point \((1,2)\) on the ellipse \(3 x^2+4 y^2=19\) is also a tangent to the parabola \(\mathbf{y}^2-\mathbf{k x}=\mathbf{0}\), then \(\mathbf{k}=\)

120642 If the tangent at a point \(P\) one the parabola \(y^2=\) \(3 x\) is parallel to the line \(x+2 y=1\) and tangents at the point \(Q\) and \(R\) on the ellipse \(\frac{x^2}{4}+\frac{y^2}{1}=1\) are perpendicular to the line \(x-y=\) 2, then the area of the triangle \(P Q R\) is:

120643 Let a circle of radius 4 be concentric to the ellipse \(15 x^2+19 y^2=285\). Then the common tangents are inclined to the minor axis of the ellipse at the angle.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-21.04.2019,Shift-I], So, Hence, Where, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), On comparing standard ellipse equation, \(\mathrm{a}^2=19, \mathrm{~b}^2=15\), Equation of tangent of ellipse is -, \(\mathrm{y}=\mathrm{mx}+\sqrt{19 \mathrm{~m}^2+15}\), \(\mathrm{mx}-\mathrm{y} \pm \sqrt{19 \mathrm{~m}^2+15}=0\), Perpendicular distance from centre of circle to tangent \(=\), 4, \(\left|\frac{ \pm \sqrt{19 \mathrm{~m}^2+15}}{\sqrt{\mathrm{m}^2+1}}\right|=4\), \(19 \mathrm{~m}^2+15=16 \mathrm{~m}^2+16\), \(\mathrm{~m}= \pm \frac{1}{\sqrt{3}}\), \(\tan \theta=\mathrm{m}=\frac{1}{\sqrt{3}}\), \(\theta=\frac{\pi}{6}\) with \(\mathrm{x}\)-axis or \(\frac{\pi}{3}\) with \(\mathrm{y}\)-axis, the common tangents are inclined to the minor axis of the ellipse at an angle of \(\frac{\pi}{3}\)., 911. If the tangent drawn to the parabola \(y^2=4 x\) at \(\left(t^2, 2 t\right)\) is the normal to the ellipse \(4 x^2+5 y^2=20\) at \((\sqrt{5} \cos \theta, 2 \sin \theta)\), then,

120644 The value of ' \(k\) ' so that the line \(y=2 x+k\) may touch the ellipse \(3 x^2+5 y^2=15\) is

120645 If the tangent at the point \((1,2)\) on the ellipse \(3 x^2+4 y^2=19\) is also a tangent to the parabola \(\mathbf{y}^2-\mathbf{k x}=\mathbf{0}\), then \(\mathbf{k}=\)