(1a)
Let $y=mx+c$ be a tangent of $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
Let $(h,k)$ be the point of intersection.
So, $k=mh+c, b^2h^2-a^2k^2=a^2b^2\implies b^2h^2-a^2(mh+c)^2=a^2b^2$
or,$h^2(b^2-a^2m^2)- 2a^2mch -a^2b^2-a^2c^2=0 $
This is a quadratic equation in $h,$ for tangency, the roots need to be same, to make the two points of intersection coincident.
$\implies (- 2a^2mc)^2=4\cdot (b^2-a^2m^2)(-a^2b^2-a^2c^2) $
$a^2m^2c^2=a^2m^2b^2+a^2m^2c^2-b^4-b^2c^2$ cancelling out $a^2$ as $a\ne0$
$0=a^2m^2b^2-b^4-b^2c^2$
$0=a^2m^2-b^2-c^2$ cancelling out $b^2$ as $b\ne0$
$\implies c^2=a^2m^2-b^2\implies c=\pm\sqrt{a^2m^2-b^2}$
(1b)
Alternatively, we can take the central conic to be $Ax^2+By^2=1$
Applying the same method we get, $ABc^2=A+m^2B$
Here $A=\frac 1{a^2},B=-\frac 1{b^2}\implies c^2=a^2m^2-b^2$
(2a)
Using calculus as André Nicolas has already done,
we find the equation of the tangent to be $$\frac {xx_1}{a^2}-\frac{yy_1}{b^2}-1=0$$
Comparing with $mx-y+c=0$ we get, $$\frac{x_1}{ma^2}=\frac{y_1}{b^2}=\frac{-1} c$$
Now $(x_1,y_1)$ lies on the given curve, so $\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$, eliminating $x_1,y_1$, we shall get the desired result.
(2b) we know , the parametric equation of the given curve is $x=a\sec t,y=b\tan t$
So, the equation of the tangent becomes $\frac{x\sec t}a-\frac{y\tan t}b-1=0$
Comparing with $mx-y+c=0$ we get, $$\frac{\sec t}{ma}=\frac{\tan t}b=\frac {-1}c$$
So, $\sec t=-\frac{ma}c,\tan t=-\frac b c$
Now use eliminate $t$.
To find the slope of the curve, or the gradient you can differentiate the function and plug in the $x$ value of the point into the derivative and that will yield the gradient at that point. So you have $y=5-x^2$ at the point $(1, 4)$.
$$y'=-2x$$
The gradient of the curve, $y=5-x^2$ at $(1,4)$ is:
$$y'(1)=-2(1)=-2$$
For the equation of the tangent line, you have the gradient $(-2)$, a point on the line $(1, 4)$, and you know that the equation of a line is $y=mx+c$. You can plug in the values that you know to find the value of $c$ and then you'll have the equation of the tangent line.
Best Answer
First consider implicit differentiation.
$$x^2= 19.8y^2 - 6.9y + 0.7$$
$$\to \frac{d}{dx} x^2 = \frac{d}{dx} (19.8y^2 - 6.9y + 0.7)$$
$$\to 2x = 2(19.8)y \frac{dy}{dx} - 6.9 \frac{dy}{dx}$$
$$\to 2x = (2(19.8)y - 6.9) \frac{dy}{dx}$$
$$\to \frac{2x}{2(19.8)y - 6.9} = \frac{dy}{dx}$$
There is some point on the graph call it $(x_1, y_1)$ s.t. the tangent line to the graph at that point passes through $(0,0.08)$.
Thus, we have the two equations:
$$x_1^2= 19.8y_1^2 - 6.9y_1 + 0.7$$
$$\frac{0.08 - y_1}{0 - x_1} = \frac{2x_1}{2(19.8)y_1 - 6.9}$$
Solve the equations* for $x_1$ and $y_1$ then find the equation of the line that passes through $(x_1, y_1)$ and $(0,0.08)$.
*You'll probably get more than one solution. If so, do you know which one to pick?