12th Standard Mathematics — Two Dimensional Analytical Geometry-II: Online Practice Test

Every member of this family already passes through \((1,5)\) and \((4,1)\): substituting either point makes both the base circle and the bracket vanish, so the condition left to impose is tangency to the \(y\)-axis.

Collecting terms, the circle is \(x^{2}+y^{2}+(4\lambda-5)x+(3\lambda-6)y+(9-19\lambda)=0\). Comparing with \(x^{2}+y^{2}+2gx+2fy+c=0\) gives \(2f=3\lambda-6\) and \(c=9-19\lambda\).

A circle touches the \(y\)-axis exactly when the chord it cuts there shrinks to a point, i.e. \(f^{2}=c\):

Hence \(\lambda=0\) or \(\lambda=-\dfrac{40}{9}\); both give genuine circles tangent to the \(y\)-axis, so the answer collects both values.

For \(\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\) the latus rectum is \(\dfrac{2b^{2}}{a}=8\), so \(b^{2}=4a\). The conjugate axis is \(2b\) and the distance between the foci is \(2c\); the condition \(2b=\tfrac{1}{2}(2c)\) gives \(2b=c\).

Using \(c^{2}=a^{2}+b^{2}\) with \(c=2b\):

Combining with \(b^{2}=4a=4\sqrt{3}\,b\) gives \(b=4\sqrt{3}\), then \(a=12\) and \(c=2b=8\sqrt{3}\). Therefore

Rewrite the circle as \(x^{2}+y^{2}-4x-8y-5=0\); its centre is \((2,4)\) and radius \(\sqrt{4+16+5}=5\).

The line \(3x-4y-m=0\) meets the circle in two distinct points precisely when the distance from the centre to the line is less than the radius:

This unfolds to \(-25

Because the circle touches the \(x\)-axis at \((1,0)\), its centre lies directly above that point, at \((1,k)\), and its radius equals \(|k|\).

Forcing the circle through \((2,3)\):

The radius is \(\dfrac{5}{3}\), so the diameter is \(2\cdot\dfrac{5}{3}=\dfrac{10}{3}\).

A second-degree equation represents a circle only when the coefficients of \(x^{2}\) and \(y^{2}\) are equal, so \(b=3\).

Substituting and dividing through by \(3\):

Here \(g=2,\ f=-3,\ c=3\), so the radius is \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+9-3}=\sqrt{10}\).

Factor each pair of lines. From \(x^{2}-8x+12=0\) we get the vertical sides \(x=2\) and \(x=6\); from \(y^{2}-14y+45=0\) we get the horizontal sides \(y=5\) and \(y=9\).

The vertical sides are \(4\) apart and so are the horizontal sides, confirming a square. The inscribed circle is centred at the centre of the square, the midpoint of the two pairs:

A normal to any circle passes through its centre. For \(x^{2}+y^{2}-2x-2y+1=0\) the centre is \((1,1)\).

To be parallel to \(2x+4y=3\) the normal must share its slope \(-\dfrac{1}{2}\). The line through \((1,1)\) with this slope is

The defining property of an ellipse is that the sum of the two focal distances of any point on it equals the length of the major axis, \(2a\).

Here \(a^{2}=16\), so \(a=4\) and \(PF_{1}+PF_{2}=2a=8\).

Two diameters of a circle always intersect at its centre. Solving \(x+y=6\) and \(x+2y=4\) simultaneously: subtracting gives \(y=-2\), then \(x=8\), so the centre is \((8,-2)\).

The radius is the distance from the centre to the given point \((6,2)\):

For \(\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\) the foci lie on the \(x\)-axis at \(\left(\pm\sqrt{a^{2}+b^{2}},\,0\right)\). The companion curve \(\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=-1\), i.e. \(\dfrac{y^{2}}{b^{2}}-\dfrac{x^{2}}{a^{2}}=1\), has foci on the \(y\)-axis at \(\left(0,\,\pm\sqrt{a^{2}+b^{2}}\right)\).

These four points form a rhombus whose diagonals both measure \(2\sqrt{a^{2}+b^{2}}\). Its area is

For \(y^{2}=4x\) we have \(a=1\), so the latus-rectum ends are \((1,2)\) and \((1,-2)\). The tangent slopes there are \(\dfrac{2a}{y_{1}}=\pm1\), hence the normal slopes are \(\mp1\).

The two normals are

Each is tangent to the circle centred \((3,-2)\), so \(r\) equals the distance from the centre to either line:

For \(y^{2}=4ax\) the slope form of the normal is \(y=mx-2am-am^{3}\). Here \(4a=12\), so \(a=3\).

The line \(x+y=k\) is \(y=-x+k\), which has slope \(m=-1\). Substituting:

Thus \(x+y=9\) and \(k=9\).

\(E_{1}\) touches its bounding rectangle at \((\pm3,0)\) and \((0,\pm2)\), so \(R\) has corners \((\pm3,\pm2)\).

Let \(E_{2}:\dfrac{x^{2}}{A^{2}}+\dfrac{y^{2}}{B^{2}}=1\). Through \((0,4)\): \(B^{2}=16\). Through the corner \((3,2)\):

Since \(16>12\) the major axis is vertical, so

Note: the worked value \(e=\tfrac{1}{2}\) is taken as the correct answer (it is option C in the ordering used here). The supplied draft key listed option 4 for Q13 — see the ReadMe flag.

Tangents parallel to \(2x-y=1\) have slope \(m=2\). For a point \((x_{1},y_{1})\) on \(\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1\), the tangent slope is \(\dfrac{4x_{1}}{9y_{1}}\). Setting it equal to \(2\) gives \(x_{1}=\dfrac{9y_{1}}{2}\).

Substituting into the hyperbola:

Taking \(y_{1}=\dfrac{1}{\sqrt{2}}\) gives \(x_{1}=\dfrac{9}{2\sqrt{2}}\), so one point of contact is \(\left(\dfrac{9}{2\sqrt{2}},\,\dfrac{1}{\sqrt{2}}\right)\).

For \(\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1\), \(c^{2}=a^{2}-b^{2}=16-9=7\), so the foci are \(\left(\pm\sqrt{7},\,0\right)\).

A circle centred at \((0,3)\) through \(\left(\sqrt{7},0\right)\) has \(r^{2}=7+9=16\). Expanding:

Circle \(T\) is centred on the \(y\)-axis and passes through the origin, so its centre is \((0,r)\) and its radius is \(r\).

External tangency with \(C\) (centre \((1,1)\), radius \(1\)) means the distance between centres equals the sum of the radii:

Squaring: \(1+1-2r+r^{2}=1+2r+r^{2}\), which collapses to \(1=4r\), so \(r=\dfrac{1}{4}\).

The distance between the foci is \(2c=6\), so \(c=3\). With \(e=\dfrac{c}{a}=\dfrac{3}{5}\) we get \(a=5\), and \(b^{2}=a^{2}-c^{2}=25-9=16\), so \(b=4\).

The quadrilateral with the axes as diagonals has vertices \((\pm5,0)\) and \((0,\pm4)\) — a rhombus with diagonals \(2a=10\) and \(2b=8\):

Using the point \((a\cos\theta,\,b\sin\theta)\), a rectangle inscribed symmetrically has area \(A=(2a\cos\theta)(2b\sin\theta)=2ab\sin2\theta\).

This is greatest when \(\sin2\theta=1\), i.e. \(\theta=\dfrac{\pi}{4}\), giving corners \(\left(\pm\dfrac{a}{\sqrt{2}},\pm\dfrac{b}{\sqrt{2}}\right)\) and

Take \(B=(0,b)\) and foci \((\pm c,0)\). The triangle \(F_{1}BF_{2}\) is right-angled at \(B\), so by Pythagoras \(F_{1}F_{2}^{2}=BF_{1}^{2}+BF_{2}^{2}\):

Since \(b^{2}=a^{2}(1-e^{2})\) and \(c^{2}=a^{2}e^{2}\), we get \(e^{2}=1-e^{2}\), so \(e=\dfrac{1}{\sqrt{2}}\).

Read the relation as the focus–directrix law. The left side is the squared distance from \((x,y)\) to the focus \((3,4)\); the right side is \(\dfrac{1}{9}\,y^{2}\), the squared distance from \((x,y)\) to the line \(y=0\) scaled by \(\dfrac{1}{9}\).

So \(PF^{2}=e^{2}\,PM^{2}\) with \(e^{2}=\dfrac{1}{9}\), giving \(e=\dfrac{1}{3}<1\) — an ellipse.

The locus of points from which a pair of mutually perpendicular tangents touch a parabola is its directrix (the director line of the parabola).

For \(y^{2}=4x\) we have \(a=1\), so the directrix is \(x=-a=-1\), i.e. the locus of \(P\) is \(x=-1\).

Tangency to the \(x\)-axis at \((3,0)\) puts the centre at \((3,k)\) with radius \(|k|\). Passing through \((1,-2)\):

The circle is \((x-3)^{2}+(y+2)^{2}=4\), i.e. \(x^{2}+y^{2}-6x+4y+9=0\). Testing the options, \((5,-2)\) gives \(25+4-30-8+9=0\), so the circle passes through \((5,-2)\).

The description is the focus–directrix definition with focus \((-2,0)\), directrix \(x=-\dfrac{9}{2}\) and eccentricity \(e=\dfrac{2}{3}<1\), so the locus is an ellipse. Forming \(PF^{2}=\dfrac{4}{9}\,PM^{2}\):

Expanding and simplifying, \(x^{2}+4x+4+y^{2}=\dfrac{4}{9}x^{2}+4x+9\), which reduces to \(\dfrac{5}{9}x^{2}+y^{2}=5\), i.e.

Writing \(16x^{2}-9y^{2}=144\) as \(\dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1\) gives \(a^{2}=9,\ b^{2}=16\). The tangency condition for \(y=mx+c\) is \(c^{2}=a^{2}m^{2}-b^{2}\):

So \(2\) and \(-2\) are the roots of \(x^{2}-(a+b)x-4=0\). Their sum is \(2+(-2)=0\), and since the sum of the roots equals \(a+b\), we get \(a+b=0\) (the product \(-4\) matches the constant term).

The centre of \(x^{2}+y^{2}-8x-4y+c=0\) is \((-g,-f)=(4,2)\), and the centre is the midpoint of every diameter.

If one end is \((11,2)\), the other end \((x,y)\) satisfies \(\left(\dfrac{11+x}{2},\dfrac{2+y}{2}\right)=(4,2)\), giving \(x=-3,\ y=2\). Hence the other end is \((-3,2)\).