The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is

Let $\alpha$ and $\beta$ be the roots of the equation $p{x}^2+qx-r=0$, where $p\ne 0$. If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{3}{4}$, then the value of $(\alpha -\beta {)}^2$ is :

Let $\mathbf{S}=\left\{x\in \mathbf{R}:(\sqrt{3}+\sqrt{2}{)}^x+(\sqrt{3}-\sqrt{2}{)}^x=10\right\}$. Then the number of elements in $\mathrm{S}$ is :

If $$\alpha ,\beta$$ are the roots of the equation, $$x^2-x-1=0$$ and $$S_n=2023{\alpha }^n+2024{\beta }^n$$, then :

Let $$\mathrm{S}$$ be the set of positive integral values of $$a$$ for which $$\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32}

Let $$\alpha ,\beta ;\alpha >\beta$$, be the roots of the equation $$x^2-\sqrt{2}x-\sqrt{3}=0$$. Let $${\mathrm{P}}_n={\alpha }^n-{\beta }^n,n\in \mathrm{N}$$. Then $$(11\sqrt{3}-10\sqrt{2}){\mathrm{P}}_{10}+(11\sqrt{2}+10){\mathrm{P}}_{11}-11{\mathrm{P}}_{12}$$ is equal to

Let $$\alpha ,\beta$$ be the roots of the equation $$x^2+2\sqrt{2}x-1=0$$. The quadratic equation, whose roots are $${\alpha }^4+{\beta }^4$$ and $$\frac{1}{10}({\alpha }^6+{\beta }^6)$$, is:

If 2 and 6 are the roots of the equation $$a{x}^2+bx+1=0$$, then the quadratic equation, whose roots are $$\frac{1}{2a+b}$$ and $$\frac{1}{6a+b}$$, is :

The sum of all the solutions of the equation $$(8{)}^{2x}-16\cdot (8{)}^x+48=0$$ is :

Let $$\alpha ,\beta$$ be the distinct roots of the equation $$x^2-\left(t^2-5t+6\right)x+1=0,t\in \mathbb{R}$$ and $$a_n={\alpha }^n+{\beta }^n$$. Then the minimum value of $$\frac{a_{2023}+a_{2025}}{a_{2024}}$$ is

For $0<\theta <\pi /2$, if the eccentricity of the hyperbola $x^2-y^2{\mathrm{cosec}}^2\theta =5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2{\mathrm{cosec}}^2\theta +y^2=5$, then the value of $\theta$ is :

Let $$e_1$$ be the eccentricity of the hyperbola $$\frac{x^2}{16}-\frac{y^2}{9}=1$$ and $$e_2$$ be the eccentricity of the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{a}>\mathrm{b}$$, which passes through the foci of the hyperbola. If $${\mathrm{e}}_1{\mathrm{e}}_2=1$$, then the length of the chord of the ellipse parallel to the $$x$$-axis and passing through $$(0,2)$$ is :

If the foci of a hyperbola are same as that of the ellipse $$\frac{x^2}{9}+\frac{y^2}{25}=1$$ and the eccentricity of the hyperbola is $$\frac{15}{8}$$ times the eccentricity of the ellipse, then the smaller focal distance of the point $$\left(\sqrt{2},\frac{14}{3}\sqrt{\frac{2}{5}}\right)$$ on the hyperbola, is equal to

Let $$P$$ be a point on the hyperbola $$H:\frac{x^2}{9}-\frac{y^2}{4}=1$$, in the first quadrant such that the area of triangle formed by $$P$$ and the two foci of $$H$$ is $$2\sqrt{13}$$. Then, the square of the distance of $$P$$ from the origin is

Let the foci of a hyperbola $$H$$ coincide with the foci of the ellipse $$E:\frac{(x-1{)}^2}{100}+\frac{(y-1{)}^2}{75}=1$$ and the eccentricity of the hyperbola $$H$$ be the reciprocal of the eccentricity of the ellipse $$E$$. If the length of the transverse axis of $$H$$ is $$\alpha$$ and the length of its conjugate axis is $$\beta$$, then $$3{\alpha }^2+2{\beta }^2$$ is equal to

Consider a hyperbola $$\mathrm{H}$$ having centre at the origin and foci on the $$\mathrm{x}$$-axis. Let $${\mathrm{C}}_1$$ be the circle touching the hyperbola $$\mathrm{H}$$ and having the centre at the origin. Let $${\mathrm{C}}_2$$ be the circle touching the hyperbola $$\mathrm{H}$$ at its vertex and having the centre at one of its foci. If areas (in sq units) of $$C_1$$ and $$C_2$$ are $$36\pi$$ and $$4\pi$$, respectively, then the length (in units) of latus rectum of $$\mathrm{H}$$ is

Let $$H:\frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$$ be the hyperbola, whose eccentricity is $$\sqrt{3}$$ and the length of the latus rectum is $$4\sqrt{3}$$. Suppose the point $$(\alpha ,6),\alpha >0$$ lies on $$H$$. If $$\beta$$ is the product of the focal distances of the point $$(\alpha ,6)$$, then $${\alpha }^2+\beta$$ is equal to

Let $$\left(5,\frac{a}{4}\right)$$ be the circumcenter of a triangle with vertices $$\mathrm{A}(a,-2),\mathrm{B}(a,6)$$ and $$C\left(\frac{a}{4},-2\right)$$. Let $$\alpha$$ denote the circumradius, $$\beta$$ denote the area and $$\gamma$$ denote the perimeter of the triangle. Then $$\alpha +\beta +\gamma$$ is

Two vertices of a triangle $$\mathrm{ABC}$$ are $$\mathrm{A}(3,-1)$$ and $$\mathrm{B}(-2,3)$$, and its orthocentre is $$\mathrm{P}(1,1)$$. If the coordinates of the point $$\mathrm{C}$$ are $$(\alpha ,\beta )$$ and the centre of the of the circle circumscribing the triangle $$\mathrm{PAB}$$ is $$(\mathrm{h},\mathrm{k})$$, then the value of $$(\alpha +\beta )+2(\mathrm{h}+\mathrm{k})$$ equals

Consider a $△ABC$ where $A(1,3,2), B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $B C$ at $D$, then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is :