A circle that touches the $x$-axis has radius equal to the $y$-coordinate of its centre (in magnitude), because the perpendicular distance from the centre to the $x$-axis is the radius.

For centre $(h,k)$ touching the $x$-axis, $r=|k|$, so the equation is

$(x-h)^2+(y-k)^2=k^2$

Similarly, a circle touching the $y$-axis has $r=|h|$ and equation $(x-h)^2+(y-k)^2=h^2$.

For example, with centre $(2,3)$ touching the $x$-axis, $r=3$ and the circle is $(x-2)^2+(y-3)^2=9$.

(a) $c^2 = a^2(1 + m^2)$

For the line $y=mx+c$ to be tangent to the circle $x^2+y^2=a^2$, the perpendicular distance from the centre $(0,0)$ to the line must equal the radius $a$:

$\frac{|c|}{\sqrt{1+m^2}}=a \implies c^2=a^2(1+m^2).$

(a) Dividing $9x^2+16y^2=144$ by $144$,

$\frac{x^2}{16}+\frac{y^2}{9}=1$

so $a^2=16$, $b^2=9$ with $a>b$ (major axis along the $x$-axis).

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt7}{4}$

(b) Let the semi-axes be $a$ (major) and $b$ (minor). Major axis $=4\times$ minor axis gives $2a=4(2b)$, so $a=4b$.

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{b^2}{16b^2}}=\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{4}$

The point $(2,1)$ only fixes the size of the ellipse ($b^2=\tfrac54$, $a^2=20$); the eccentricity depends only on the ratio of the axes.

Let the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$. Given latus rectum $\dfrac{2b^2}{a}=5$ and $e=\dfrac12$.

Since $b^2=a^2(1-e^2)=a^2\left(1-\tfrac14\right)=\tfrac34a^2$,

$\frac{2}{a}\cdot\frac{3a^2}{4}=5\;\Rightarrow\;\frac{3a}{2}=5\;\Rightarrow\;a=\frac{10}{3}$

$b^2=\frac34\cdot\frac{100}{9}=\frac{25}{3}$

Hence, with the major axis along the $x$-axis,

$\frac{x^2}{100/9}+\frac{y^2}{25/3}=1\quad\text{i.e.}\quad 9x^2+12y^2=100$

If the major axis is along the $y$-axis, the equation is $12x^2+9y^2=100$.

The equation is missing from the question; taking the hyperbola $9x^2-16y^2=144$ (as in the related question),

$\frac{x^2}{16}-\frac{y^2}{9}=1,\qquad a^2=16,\ b^2=9$

$c^2=a^2+b^2=25\;\Rightarrow\;c=5$

Eccentricity: $e=\dfrac{c}{a}=\dfrac54$.

Foci: $(\pm c,0)=(\pm5,0)$.

Condition for the Line $y = mx + c$ to Touch the Parabola $y^2 = 4ax$

Given

• Parabola: $y^2 = 4ax$
• Line: $y = mx + c$

Step 1: Substitute the line into the parabola

From $y = mx + c$, we have $y$ in terms of $x$. Substituting into $y^2 = 4ax$:

$(mx + c)^2 = 4ax$

$m^2x^2 + 2mcx + c^2 = 4ax$

$m^2x^2 + (2mc - 4a)x + c^2 = 0$

Step 2: Apply the tangency condition

For the line to touch the parabola, it must meet it at exactly one point. So the quadratic in $x$ must have equal roots, i.e. discriminant $= 0$:

$(2mc - 4a)^2 - 4(m^2)(c^2) = 0$

$4(mc - 2a)^2 - 4m^2c^2 = 0$

$(mc - 2a)^2 - m^2c^2 = 0$

$m^2c^2 - 4amc + 4a^2 - m^2c^2 = 0$

$-4amc + 4a^2 = 0$

$4a(a - mc) = 0$

Since $a \neq 0$:

$\boxed{c = \frac{a}{m}}$

Comment on the relation of $a$, $c$ and $m$

The condition for the line $y = mx + c$ to touch the parabola $y^2 = 4ax$ is:

$c = \frac{a}{m} \quad \text{or equivalently} \quad mc = a$

• The y-intercept $c$ depends on both the slope $m$ and the focal parameter $a$.
• The product $mc$ equals $a$ — a fixed constant for a given parabola.
• For a steeper line (larger $m$), the intercept $c$ must be smaller, and vice versa, so that $mc = a$ always holds.
• This relation is independent of the point of contact; it guarantees tangency for any slope $m \neq 0$.

(b) $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

This is a hyperbola, and only a hyperbola has two axes — the transverse axis (along which the two branches lie, here the $x$-axis, of length $2a$) and the conjugate axis (the perpendicular axis, here the $y$-axis, of length $2b$). The ellipse in (a) has major and minor axes, the parabola in (c) has only one axis, and the circle in (d) has no such axes.

For the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$,

$a^2=25,\ b^2=9$, so $a=5,\ b=3$.

$c^2=a^2-b^2=25-9=16 \Rightarrow c=4$

Eccentricity: $e=\frac{c}{a}=\frac{4}{5}$

Foci: $(\pm c,0)=(\pm 4,0)$.

Answer: Eccentricity $=\dfrac45$, foci $=(4,0)$ and $(-4,0)$.

For the circle $x^2+y^2=25$, centre $=(0,0)$, radius $=5$.

At the point $(3,4)$:

Tangent:
For $x^2+y^2=a^2$, tangent at $(x_1,y_1)$ is
$xx_1+yy_1=a^2$ So, $3x+4y=25$

Normal:
The normal passes through the centre $(0,0)$ and the point $(3,4)$.
Its slope is $m=\frac{4-0}{3-0}=\frac43$ Thus, $y-0=\frac43(x-0)$ $4x-3y=0$

Answer:
Tangent: $\boxed{3x+4y=25}$
Normal: $\boxed{4x-3y=0}$

Let $P(x,y)$ be any point on the parabola. Focus $F(2,3)$, directrix $x=4$. By definition, $PF=$ distance of $P$ from the directrix:

$\sqrt{(x-2)^2+(y-3)^2}=|x-4|$

Squaring,

$(x-2)^2+(y-3)^2=(x-4)^2$

$x^2-4x+4+(y-3)^2=x^2-8x+16$

$(y-3)^2=12-4x$

$\boxed{(y-3)^2=-4(x-3)}$

(b) Ellipse

Complete the square in $x$:

$4(x^2-4x)+y^2=48\;\Rightarrow\;4(x-2)^2-16+y^2=48\;\Rightarrow\;4(x-2)^2+y^2=64$

$\frac{(x-2)^2}{16}+\frac{y^2}{64}=1$

Both squared terms are positive with unequal denominators, so the conic is an ellipse (centre $(2,0)$, major axis along the $y$-axis).

$9x^2-16y^2-144=0\;\Rightarrow\;\frac{x^2}{16}-\frac{y^2}{9}=1$

This has the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ with $a^2=16$, $b^2=9$.

Centre: $(0,0)$.

Foci: $c^2=a^2+b^2=16+9=25$, so $c=5$. The transverse axis lies along the $x$-axis, so the foci are $(\pm5,0)$.

The standard equation of a circle with centre $(h, k)$ and radius $r$ is

$(x - h)^2 + (y - k)^2 = r^2$

When the centre is the origin $(0,0)$, it reduces to $x^2 + y^2 = r^2$.

The tangent to $x^2+y^2=a^2$ at $(x_1,y_1)$ is $xx_1+yy_1=a^2$.

Here $a^2=25$ and $(x_1,y_1)=(3,4)$ (which lies on the circle, as $9+16=25$), so

$\boxed{3x+4y=25}$

For an ellipse centred at the origin with foci $(\pm c,0)$,

$c=2,\qquad e=\frac12$

Since $e=\dfrac ca$,

$a=\frac ce=\frac{2}{1/2}=4$

Now,

$b^2=a^2-c^2=4^2-2^2=16-4=12$

Hence the equation of the ellipse is

$\boxed{\frac{x^2}{16}+\frac{y^2}{12}=1}$

(d) ellipse

A point $P$ with $PF_1+PF_2=\text{constant}$ (greater than $F_1F_2$) traces an ellipse with foci $F_1$ and $F_2$. This is the focal-distance definition of an ellipse.

(b) $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

The eccentricity expression appears mistyped; the standard eccentricity of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b$) is

$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{a^2-b^2}}{a}$

which has the form of the given expression with $\sqrt{\ }$ over $a^2-b^2$. The hyperbola (a) has $e=\dfrac{\sqrt{a^2+b^2}}{a}$, and the circle (c) has $e=0$, so the conic is the ellipse.